Guide :

# WRITE AN EXPRESSION THAT MEANS -1.25/1 TIMES THE RACE NUMBER MINUS 3.5

## Research, Knowledge and Information :

### Writing Expressions - BIG IDEAS MATH

x + 14 The phrase “more than” means addition. b. A number y minus 75 ... 6 times 50 3. 25 less than a number b ... Section 1.2 Writing Expressions 13 Write the ...

### Writing Expressions (solutions, examples, videos)

Writing Expressions. ... Expression . 6 more than 5 times a number . 5x + 6 . ... 1: Half of a number is 16. Write an equation to represent the situation.

### Writing Algebraic Expressions

Writing Algebraic Expressions is presnted by Math Goodies. Learn how to translate verbal phrases into algebraic expressions. ... three times a number decreased by ...

### Algebra - Math League

To evaluate an expression at some number means we replace a ... second expression for "four times my age, minus ... the list is the number 1. 1, 2, 3, 4, 5, 6 ...

### Numerical Expression - Math Dictionary - ICoachMath.com

... what is an numerical expression ... means 'phrase'. A numerical expression is a ... number. For example, the numerical expression ...

### Algebraic expressions - A complete course in algebra

... x plus 5 times x minus 2? (x + 5) ... To "evaluate" means to name and write a number. Example 1. ... 3x + 1 . Now an algebraic expression is not a sentence, ...

### Writing Expressions and Equations

Lesson 7.1 Writing Expressions and Equations 319 Write the verbal phrase ... 15. 13 is equal to 5 minus a number. 16. 1 4 ... Lesson 7.1 Writing Expressions and ...

### Name: Unit3 #2 Writing Expressions and Equations

Writing Expressions and Equations ... The ratio of z and 3 Practice: Underline key words. Write each phrase as an algebraic ... is 5 times a product of 5 and a number

### Translating Words into Mathematical Symbols

Translating Words into Mathematical Symbols ... In the expression 2x + 3 ... “To factor” means to write a number as the product of its factors.

## Suggested Questions And Answer :

### WRITE AN EXPRESSION THAT MEANS -1.25/1 TIMES THE RACE NUMBER MINUS 3.5

maebee yu meen -1.25*(x-3.5) ??????

### write a system of two linear equations showing the distance of each animal can travel to model a fair race.

One way to resolve this is to work out the distance each must run to take the same time to complete the race. Distance=speed times time, so time=distance÷speed. So if the distances are D1 and D2 we can write D1/43=D2/35 (time for coyote=time for rabbit) and that means D1/D2=43/35 (about 1.23). If the ratio of the distances is the inverse of the speeds then the race will be fair. (D1=43T (coyote) and D2=35T (rabbit); divide one equation by the other: D1/D2=43T/35T so D1/D2=43/35.) TRACK ARRANGEMENTS FOR FAIR RACE We could have two parallel linear tracks but the rabbit starts at a different point along its track. If the longer track is 43n miles long then the shorter track is 35n miles long where n is a number, probably a fraction, for example 1/10. The rabbit's start-line is 43n-35n=8n miles beyond the coyote's. The rabbit runs only 35n miles while the coyote runs 43n miles. The finishing-lines are in line. The race will take n hours if both animals run at top speed on average throughout the race. So if, for example, n=1/10 the race will last for 1/10 hr = 6 minutes=4.3/43=3.5/35. The rabbit runs 3.5 miles while the coyote runs 4.3 miles. Another way is to use circular tracks. Since the circumference of a circle is a fixed multiple (2π) of the radius, we have two concentric circular tracks where the radius of the inner track is R1 (for the rabbit) and radius of the outer track is R2 so that R1/R2=35/43=2πR1/2πR2, where 2πR1 and 2πR2 are the lengths of the inner and outer tracks. The advantage of circular tracks is that the race can be run over a number of laps rather than just one lap. The race is more exciting because spectators can surround the track and observe the progress of the race better. A third solution is to have semicircular tracks placed at the two ends of a rectangular circuit. The curvature of the circles compensates for the difference in speeds of the animals. The width of the rectangle is 2R1, where R1 is the radius of the inner track. If L=length of the rectangle then the length of the tracks are (2L+2πR1) and (2L+2πR2), and (2L+2πR1)/(2L+2πR2)=35/43=(L+πR1)/(L+πR2), so 43L+43πR1=35L+35πR2; 8L=π(35R2-43R1) (35R2>43R1, so R1/R2<35/43). This equation relates the length of the rectangular part of the track to the inner and outer radii of the semicircular parts of the tracks. If this equation is satisfied the start and finish lines will be in line. If the equation is not satisfied then, the start and finish lines will be out of line by the value of the expression 8L-π(35R2-43R1). (EXAMPLE: If R2=2R1 for the above type of track, 8L=π(70R1-43R1)=27πR1, and L=27πR1/8=10.6R1 approx. or 5.3W where W=2R1 is the width of the rectangle.) This answer probably goes into more detail than you require, but I hope you get the general idea.

Question: find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4 ? Let us rewrite your problem statement. find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4, or v1, v2, v3, v4, v5 Turning those statements above into mathematical expressions, 3 times the product of the 3rd and 4th,                            becomes   3*v3*v4 minus the product of the 2nd and (4 less than the first),  becomes   - v2*(v1-4) plus 3 times the 5th                                                          becomes   + 3*v5 is 232.                                                                               becomes   = 232 As a single expression, this is 3*v3*v4 - v2*(v1 - 4) + 3*v5 = 232 3(x+2)(x+3) - (x+1)(x - 4) + 3(x+4) = 232 3(x^2 + 5x + 6) - (x^2 - 3x - 4) + (3x + 12) = 232 2x^2 + 21x + 34 = 232 2x^2 + 21x - 198 = 0 (2x + 33)(x - 6) = 0 x = -33/2, x = 6 But x is an integer, so the only solution is x = 6 The numbers then are : 6, 7, 8, 9, 10

### Creata a fraction story

FIVE LITTLE PIGS GO TO THE MALL Mama Pig gave her five little pigs seven and a half dollars between them to spend at the mall. It was a cold day, twenty-three Fahrenheit, minus five Celsius, or five degrees below freezing. Off they trotted at a quarter to three in the afternoon. "How far is it?" the youngest pig asked after a while. "One point seven five miles from home," said the eldest. "What does that mean?" asked the youngest. "Well," explained the eldest, "if we divide the distance into quarter miles, it's seven quarters." "How long will it take to get there?" asked the pig in the middle. Her twin sister replied, "It's five past three now, so that means we've taken twenty minutes to get here. Remember the milestone outside our house? There's another one here, so we've come just one mile and we've taken a third of an hour. [That means our speed must be three miles an hour.] "How much longer?" the youngest asked. "Three quarters of a mile to go," the next youngest started. "Yes," said the eldest, "we get the time by dividing distance by speed, so that means three quarters divided by three, which is one quarter of an hour [which is fifteen minutes]." ["So what time will we arrive?" the youngest asked. "About twenty past three," all the other pigs replied together.] "That's thirty-five minutes altogether," the eldest continued, "which means that - let me see - seven quarters divided by three is seven twelfths of an hour. One twelfth of an hour is five minutes [so seven twelfths is thirty-five minutes]. Yes, that's right." When they got to the mall, it had started to snow. Outside there was a big thermometer and a sign: "COME ON IN. IT'S WARMER INSIDE!" The eldest observed: "It shows temperature in Fahrenheit and Celsius. See, there's a scale on each side of the gauge. It's warmer now than it was when we left home. The scales are divided into tenths of a degree. It says twenty-seven Fahrenheit exactly, and, look, that's the same as minus two point eight Celsius," speaking directly to the youngest, "because the top of the liquid is about eight divisions between minus two and minus three. Our outside thermometer at home is digital, but this is analogue." The youngest stuttered: "What's 'digital' and 'analogue'?" "Well," began one of the twins, "your watch is analogue, because it has fingers that move round the watch face. Our thermometer at home is digital, because it just shows the temperature in numbers." ["Yes," said the eldest, "it would show thirty-seven point zero degrees Fahrenheit, four degrees warmer, and minus two point eight Celsius, two point two degrees warmer than when we left."] Continued in comment...

### how do you solve 3t^4+2t^3-300t-50=0

The expression does not factorise and the roots seem to be t=-0.16669 and 4.48653 approx. Check that you haven't missed out a term (e.g., t^2). If the equation was meant to factorise completely, then the product of the factors=-50. The factors must therefore consist of 1, 2, 5, 5. The factors of the 3t^4 must consist of 3t, t, t, t. The minus sign in front of the constant 50 implies that there must be an odd number of pluses and minuses amongst the factors: one minus and 3 pluses, or three minuses and one plus.  Let the coefficient of the apparently missing t^2 be k. The equation then reads: 3t^4+2t^3+kt^2-300t-50=0. If (t-1) is a factor then 3+2+k-300-50=0 and k=345. If (t+1) is a factor then 3-2+k+300-50=0 and k=52-303=-251. If (t-2) is a factor then 48+16+4k-600-50 and k=(1/4)(650-64)=293/2. If (t+2) is a factor then 48-16+4k+600-50 and k=(1/4)(66-648)=-291/2. It seems improbable that (t-2) or (t+2) could be a factor because k would be a mixed number (or improper fraction), so k=345 or -251. If (t-5) is a factor then 1875+250+25k-1500-50=0 and k=(1550-2125)/25=-23. If (t+5) is a factor then 1875-250+25k+1500-50=0 and k=(300-3375)/25=-123. We know that (3t+f) must be included as a factor, but if f=-5 or +5 then, since we need two 5's within the set of factors, so if one contains 3t, the other can only contain t. That means that (t-5) or (t+5) must be a factor. There's no point trying other factors like (t-25) or (t+10), because they would require (t-2) or (t+2), or (t-5) or (t+5), and (t-1) and (t+1) as other factors, which have already been covered. The exercise shows that no one value of k allows us to find more than one root. Had any two values of k above been the same we would have had at least two roots and a quadratic, which could have been solved, if it had had real roots. The conclusion is that finding one root only leaves us with a cubic equation that has no rational, real roots. This means that the question has been wrongly stated, and the inclusion of kt^2 doesn't help to solve it satisfactorily.

### k^2-k-20/18

I read this as (k^2-k-20)/18. What pair of factors of 20 differ by 1? This is essentially what the quadratic in k is asking. The pairs of factors of 20 are (1,20), (2,10), (4,5) and the only pair with a difference of 1 is (4,5). The larger of the two has a minus and the other a plus to make the difference -1 which is the coefficient of the middle term. So k^2-k-20 factorises: (k-5)(k+4). And we place this over 18: (k-5)(k+4)/18. This won't factorise any further. When I was reviewing this question I felt there was more to it. Supposing the question was: what values can k take so that the expression is an integer? I spotted something interesting. The zeroes of the quadratic are 5 and -4: either of these values make the expression zero. But 5+4=9, and 9 is a factor of 18, the denominator. We can write the factors of the quadratic as y(y-9)/18, where y replaces k+4; or we can write y(y+9)/18, where y replaces k-5. The pairs of factors of 18 are (1,18), (2,9), (3,6). If the numerator contains any of these pairs, the expression will be an integer, or whole number, positive or negative. What values of y must we have so that y(y-9)/18 is an integer? Let's start with y=0; the expression is zero, which is an integer. Now y=3; we have 3(-6)/18=-1, another integer. y=6; 6(-3)/18=-1. y=9; we get 0. y=12 we get 12*3/18=2. y=15; we get 15*6/18=5. In fact, all multiples of 3 work, so we can write y=3N as the general solution, where N is a positive or negative integer. What about k? We know that y=k+4 or k-5 so k+4=3N or k-5=3N are solutions. We can write these solutions as k=3N-4 or k=3N+5. If N=0 we have -4 and 5, which are the zeroes of the quadratic. If N=1, k=-1 or 8; if N=2, k=2 or 11; if N=3, k=5 or 14; if N=4, k=8 or 17; and so on. If you substitute these and other values of k (according to the formula) you'll see that the expression is always an integer.