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2 years ago

Does the equation y=-2x-3 represent a linear function

Mathematics

2 answers:

Svetllana [295]2 years ago

8 0

Answer:

For example, the linear equation 3y + 2x = 6 has an x intercept when y = 0, so 3(0) + 2x = 6. The x-intercept is (3, 0). Likewise the y-intercept occurs when x = 0. The y-intercept is (0, 2).

...

Graph the linear equation y + 3x = 5.

x values 5 – 3x

Step-by-step explanation:

hope this helps

Marysya12 [62]2 years ago

3 0

Answer:

yes

Step-by-step explanation:

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What is the equation 22-2h=-5?

Marrrta [24]

Answer:

h = -9

Step-by-step explanation:

Simplifying

5h + 22 + -2h = -5

Reorder the terms:

22 + 5h + -2h = -5

Combine like terms: 5h + -2h = 3h

22 + 3h = -5

Solving

22 + 3h = -5

Solving for variable 'h'.

Move all terms containing h to the left, all other terms to the right.

Add '-22' to each side of the equation.

22 + -22 + 3h = -5 + -22

Combine like terms: 22 + -22 = 0

0 + 3h = -5 + -22

3h = -5 + -22

Combine like terms: -5 + -22 = -27

3h = -27

Divide each side by '3'.

h = -9

Simplifying

h = -9

4 0

2 years ago

A reel of electrical cable holds 80 metres when full.

kaheart [24]

I3/I6(8O)= 65m of electrical cable left

6 0

2 years ago

¼ x 20<br><br> ⅛ x 56<br><br> 0.16 x 32<br><br> 4.5 x 9.1

Kay [80]

Answer:

¼ x 20 = 5

⅛ x 56 = 7

0.16 x 32 = 5.12

4.5 x 9.1 = 40.95

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

“baka”<br> ,killua zoldyc

adell [148]

What is the question??

4 0

3 years ago

Read 2 more answers

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

Leokris [45]

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

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 ~

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

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

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

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

・．━━━━━━━†━━━━━━━━━．・

<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 ~

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

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

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

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

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<h3>Problem 3 </h3>

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

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

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

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

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<h3>Problem 4</h3>

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

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

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

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

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

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

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

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

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8 0

2 years ago