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2 years ago

20 points , Please help !!

Mathematics

2 answers:

dezoksy [38]2 years ago

7 0

Answer:

Answer A

Step-by-step explanation:

Fudgin [204]2 years ago

5 0

Answer:

y = -1/4*x + b

Insert the Point and solve for b

3 = -1/4*8 + b

3 = -2 + b

5 = b

Answer A is correct.

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Help please! any1 know this

FrozenT [24]

The area of square: s · s = s²

The area of rectangle: (s + 6) · s = s(s + 6)

So:

s² + s(s + 6) = 56

s² + s(s + 6) - 56 = 0

Answer D.

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2 years ago

Compare 3.5 • 10^4 to standard form

GREYUIT [131]

Answer:

35,000

Step-by-step explanation:

^4 means 4 zeros

10^4 = 10,000

3.5 times 10,000 =

35,000

7 0

1 year ago

Write three different equations that have x=5 as a solution

svlad2 [7]

Answer: The three different equations will be

$x+3=8\\\\2x-8=2\\\\4x-4=16$

Step-by-step explanation:

We have to write three equation that have x = 5 as a solution:

1) First equation will be

$x+3=8\\\\x=8-3\\\\x=5$

2) Second equation will be

$2x-8=2\\\\2x=8+2\\\\2x=10\\\\x=\frac{10}{2}\\\\x=5$

3) Third equation will be

$4x-4=16\\\\4x=16+4\\\\4x=20\\\\x=\frac{20}{4}\\\\x=5$

Hence, the three different equations will be

$x+3=8\\\\2x-8=2\\\\4x-4=16$

7 0

3 years ago

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i have 537 cookies and i need to divide them into 12 equal groups. how many cookies are in each group

katrin [286]

If you divide 537 in to 12 groups. each group would have 44 cookies and you would have 9 left.

5 0

2 years ago

Write the expression 44(4–7)(4) using a single exponent.

Ghella [55]

<h3>The given expression as single exponent is:</h3>

$4^4 \times 4^{-7} \times 4 = 4^{-2}$

<em><u>Solution:</u></em>

<em><u>Given expression is:</u></em>

$4^4 \times 4^{-7} \times 4$

In exponents,

When the base is same, exponents can be added

Which means,

$a^m \times a^n = a^{m+n}$

Therefore,

$4^4 \times 4^{-7} \times 4 = 4^{4-7+1}\\\\Simplify\\\\4^4 \times 4^{-7} \times 4 = 4^{-3+1} \\\\4^4 \times 4^{-7} \times 4 = 4^{-2}$

Thus the given expression as single exponent is:

$4^4 \times 4^{-7} \times 4 = 4^{-2}$

8 0

2 years ago

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