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FromTheMoon [43]

3 years ago

8

Shorts are on sale for 10% off. If Kate buys 10 pairs of shorts at s dollars apiece, which expression is NOT equivalent to the s

ituation stated?
A) 9s
B) 10s − 0.10(10s)
C) 10s − 0.10s
D) 10s − 1s

2 answers:

NeX [460]3 years ago

7 0

Answer:

The correct answer is C

Step-by-step explanation:

SIZIF [17.4K]3 years ago

3 0

10 pairs of short cost 10s (without sale).

10% =0.1, and 10% off is

10s - 0.10*10s (B) = 10s -1s (D)

10s -1s = 9s (A)

Expression that does not match is (C),

correct answer is (C).

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Write the equation in function form (solved for y alone) 2y - 8x = 6.

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Answer:

Move all terms that don't contain y to the right side and solve.

y = 3 + 4 x

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Directions: Calculate the area of a circle using 3.14x the radius

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$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

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<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

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<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

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$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

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・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

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