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3 years ago

Count by tens and ones to find sum use a number line 36+28

Mathematics

1 answer:

finlep [7]3 years ago

4 0

30+20+6+8
=64

(not enought word count so im going to type some random stuffjnfdjjkldfja)

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Translation : 2 left and 4 down

Marizza181 [45]

This is the correct answer to the problem

3 0

3 years ago

Read 2 more answers

WARNJNG ⚠️⚠️ just help me please

marta [7]

(4)3 is also the same thing as 4x3 which is 12n

5 0

2 years ago

Find values a and b that satisfy a∙〈5,-4,0〉+b∙〈2,2,-3〉=〈11,-16,-6〉.

WARRIOR [948]

The right hand side of the equation is 〈11 , -16 , 6〉

The values of a and b are a = 3 , b = -2

Step-by-step explanation:

If n . <a , b> + m . <c , d> = < x , y>, then

na + mc = x
nb + md = y

∵ a ∙〈5 , -4 , 0〉 + b ∙ 〈2 , 2 , -3〉 = 〈11 , -16 , 6〉

- By using the rule above

∴ a(5) + b(2) = 11 ⇒ (1)

∴ a(-4) + b(2) = -16 ⇒ (2)

∴ a(0) + b(-3) = 6 ⇒ (3)

Use equation (3) to find the value of b

∵ a(0) + b(-3) = 6

∴ 0 - 3b = 6

∴ -3b = 6

- Divide both sides by -3

∴ b = -2

Substitute b in equation (1) or equation (2) to find a

∵ a(5) + b(-2) = 11

∴ 5a - 2b = 11

∵ b = 2

∴ 5a - 2(2) = 11

∴ 5a - 4 = 11

- Add 4 to both sides

∴ 5a = 15

- Divide both sides by 5

∴ a = 3

To check your answer substitute a and b in equation (2)

∵ a(-4) + b(2) = -16

∴ -4a + 2b = -16

∵ a = 3 , b = -2

∵ The left hand side is -4a + 2b = -4(3) + 2(-2) = -12 - 4 = -16

∵ The right hand side is -16

∴ L.H.S = R.H.S

∴ The values of a and b are 3 , -2

The values of a and b are a = 3 , b = -2

Learn more:

You can learn more about the equations in brainly.com/question/11306893

#LearnwithBrainly

8 0

3 years ago

Simplify 45÷2×8-12+42 Please show working

Allushta [10]

Answer:

210

Step-by-step explanation:

Lets use BEDMAS (Brackets, Exponents, Division, Multiplication, Addition, Subtraction)

45 ÷ 2 = 22.5

22.5 × 8 = 180

180 - 12 = 168

168 + 42 = 210

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Have a great summer :)

6 0

3 years ago

Please help! Correct answer only, please!

sukhopar [10]

Answer: A. $\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]$

Step-by-step explanation:

The question is asking us to find the product of the matrices. The key difference is the second A has a little T in the exponent. This T means transpose. You multiply A by the transpose of A. To find the transpose, you turn the rows into columns.

$A^T=\left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right]$

Now that we have our transpose, we can multiply the matrices.

$\left[\begin{array}{ccc}5&2\\3&-1\\\end{array}\right] \left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right] =\left[\begin{array}{ccc}5*5+2*2&5*3+2(-1)\\3*5+2(-1)&3*3+(-1)(-1)\\\end{array}\right] =\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]$

4 0

3 years ago