Guide :

# How can 10 × 4 help you solve 9 × 4? Explain.

How can 10 × 4 help you solve 9 × 4? Explain.

## Research, Knowledge and Information :

### Make 10 to add - Langford Math homepage

[the teacher invites a child to explain how they made a 10 to solve ... make 10, they can sometimes use them to solve ... 6+4=10, then they can figure ...

### Use Addition to Solve Subtraction - CPALMS.org

What added to seven makes 10? How can we use that to help us know ... The student can explain the ... Do you think we can always use addition to solve subtraction ...

### Equations may vary. Order of Write two equations for each ...

Write two equations for each Math Mountain. 4. ... Explain how you can make a ten to find the partner. ... drawing can help you solve 8 + = 14. 13 8 15 9 13

### Developing the Concept - eduplace.com

Developing the Concept. ... You can use doubles to help you ... you can make a 10 by taking 1 from the 5 and adding it to the 9 to make 10. Now you have a 10 and a 4 ...

### Using Addition to Solve Subtraction Problems

The student can explain the ... the student cannot identify an addition equation that could help solve ... Will this always work with subtraction problems? Can you ...

### Math Connects 5th grade Essential Questions

th grade: Essential Questions 4-1 Explain how placing zeros at the ... 9-6 How can looking for a pattern help you solve problems? 9-7 Will multiplying two numbers ...

### Steps to Solving Equations - svmimac.org

Steps to Solving Equations ... This lesson unit is intended to help you assess how well students are able to: ... Make sure you explain your answers really

### WebMath - Solve Your Math Problem

WebMath is designed to help you solve your math problems. ... Webmath is a math-help web site that generates answers to specific math questions and problems, as ...

### Write two equations for each Math Mountain. - PBworks

Write two equations for each Math Mountain. 4. ... Explain how you can make a ten to find the partner. ... drawing can help you solve 8 + = 14. 13 8 15 9 13 6

## Suggested Questions And Answer :

please help me solve 1/2(x+1)=1/3(x-1) please help me solve thie equation showing work 1/2(x + 1) = 1/3(x - 1) 6(1/2(x + 1)) = 6(1/3(x - 1)) 3(x + 1) = 2(x - 1) 3x + 3 = 2x - 2 x = -5 Check it. 1/2(x + 1) = 1/3(x - 1) 1/2(-5 + 1) = 1/3(-5 - 1) 1/2(-4) = 1/3(-6) -2 = -2

Problem: (-3x^2)^3= Please help me solve please help me solve (-3x^2)^3 (-3x^2)^3 = (-3x^2)(-3x^2)(-3x^2) Solve it piecemeal: -3 * -3 = 9 9 * -3 = -27 x^2 * x^2 = x^4 x^4 * x^2 = x^6 Putting them together, (-3x^2)^3 = -27x^6

### How do you solve this problem? solve for d: n= m+3d/4

Problem: How do you solve this problem? solve for d: n= m+3d/4 I need help solving this and seeing how it was solved since I am having trouble figuring out what to do and how to solve it thanks. n = m + 3d/4 n - m = (m + 3d/4) - m n - m = 3d/4 4(n - m) = (3d/4)4 4(n - m) = 3d 4(n - m)/3 = 3d/3 4/3 (n - m) = d Answer: d = 4/3 (n - m)

### -4d + 2e + f= -20

-4d + 2e + f = -20 pleasee help meeeee There is nothing you can do with this. When you have more than one unknown (one variable), you need more than one equation. You manipulate the equations by multiplying them so two of them have the same co-efficient for one of the variables. Then, you add or subtract those equations to eliminate the variable. Keep doing that until you have one equation with only one variable. Solve that equation and plug the value into one of the equations to solve for another variable. Keep adding and subtracting equations to isolate one of the variables, solve for it, and use that to help solve another equation. One equation with two or more variables is a lost cause.

### How to solve (3x+22)=(6x-8)?

How to solve (3x+22)=(6x-8)? Our lesson is now about Parallel Lines and we have to solve first before using the theorems. And I don't know how to solve it. Please help me. Oh, and the teacher said thar the working equations is m<1 = m<2. Thank you so much 3x + 22 = 6x - 8 3x + 22 - 3x = 6x - 8 - 3x 22 = 3x - 8 22 + 8 = 3x - 8 + 8 30 = 3x 30 / 3 = 3x / 3 10 = x

### using the elimination method solve 6=6x+6y and -14y=18x+14

using the elimination method solve 6=6x+6y and -14y=18x+14 help me solve step by step 1) 6 = 6x + 6y 2) -14y = 18x + 14 Re-write equation 1: 3) 6x + 6y = 6 It is standard to have the unknowns of the left side of the equation. Now, for equation 2. -14y = 18x + 14 4) -18x - 14y = 14 Multiply equation 3 by 3. 3 * (6x + 6y) = 6 * 3 5) 18x + 18y = 18 Add equation 4 to equation 5.    18x + 18y = 18 +(-18x - 14y = 14) -------------------           4y = 32 You have just eliminated the x. Solve for y. 4y = 32 y = 8 You can used either equation 1 or equation 2 to solve for x. 1) 6 = 6x + 6y = 6    6x + 6(8) = 6    6x + 48 = 6    6x = 6 - 48    6x = -42    x = -7 Check it with equation 2. 2) -14y = 18x + 14    -14(8) = 18(-7) + 14    -112 = -126 + 14    -112 = -112 You have x = -7 and y = 8

### need help in solving for x ?

2x - 7 = -x + 2 add 2 to both sides: 3x - 7 = 2 add 7 to both sides: 3x = 9 divide both sides by 3: x = 9/3 = 3   -2 - 3x = x + 6 subtract x from both sides: -2 - 4x = 6 add 2 to both sides: -4x = 8 divide both sides by -4: x = 8/-4 = -2

### Help needed in understanding the process to solve a trig problem such as [ 2sin^2x-3x= 0 ]

Remember the relation between sine and cosine? sine^2+cosine^2=1, so sin^2X=1-cos^2X. Substitute for sin^2X: 2(1-cos^2X)-3cosX=0; 2-2cos^2X-3cosX=0. We now have a quadratic equation. If it helps, let y= cosX, then: 2-2y^2-3y=0 or 2y^2+3y-2=0, which factorises: (2y-1)(y+2)=0, and y=1/2 or -2 which means cosX=1/2 or -2. But we know that cosX can't be bigger than 1 or less than -1, so we can safely reject cosX=-2, and solve cosX=1/2. This is one of those familiar angles, X=60 degrees or (pi)/3 radians. Now you need to remember which quadrants cosine is positive in. ASTC (All Silver Tea Cups: All trigs, sin, tan, cos) for Q1 to 4. So cosine is positive in Q1 and Q4. The solution therefore is (pi)/3 and 2(pi)-(pi)/3=5(pi)/3 or 60 and 360-60=300 degrees.