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3 years ago

ASAP:( Match each system of equations to the diagram that represents its solution.

Mathematics

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:i need this answer

Step-by-step explanation:

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Use a scale of 2km to 1cm to find the length of a road

noname [10]

<h2>First Question:</h2><h3>Given: </h3>

2 km = 1 cm

The length of the road of 15 km.

<h3>Solution:</h3>

2 km = 1 cm

1 km = 1/2 = 0.5 cm

15 km = 0.5 × 15 = 7.5

<h2>Answer: </h2>

The length of the road is <u>7.5 cm</u>.

Therefore, <u>Option D. 7.5 cm</u> is correct.

8 0

3 years ago

What is the slope of the line that passes through the pair of points (1/3, -1) and ( -2, 9/2)

UkoKoshka [18]

Answer:

$\boxed{m = - \frac{33}{14} }$

Step-by-step explanation:

Use the form below

$\boxed{\boxed{m = \frac{y_2 - y_1}{x_2 - x_1} }}$

Where

$m$ is a slope
$(x_1,~y_1)$ and $(x_2,~y_2)$ are the point of the line

So, the slope is

$(\frac{1}{3} ,~ -1) \to x_1 = \frac{1}{3} ~and~y_1 = -1$

$(-2,~ \frac{9}{2} ) \to x_2 = -2~and~y_2 = \frac{9}{2}$

$m = \frac{\frac{9}{2} -(-1)}{-2-\frac{1}{3} }$

$m = \frac{\frac{11}{2} }{\frac{-7}{3} }$

$m = \frac{11}{2} \div (-\frac{7}{3} )$

$m = \frac{11}{2} \times (-\frac{3}{7} )$

$m = - \frac{33}{14}$

Happy to help:)

7 0

3 years ago

Read 2 more answers

Trevor Sherr is paid $1.20 for each hand-painted

Margarita [4]

Answer:

$112.80

Step-by-step explanation:

First, add 56 and 44, which is 100. Then because 6 of the dishes contained errors, he didn't sell them, so subtract 6 from 100. You will get 94. Then you multiply 94 by 1.20 which would give you a result of $112.80.

8 0

3 years ago

What is the surface area of this cuboid?<br> height: 5cm<br> length: 9cm<br> width: 4cm

bazaltina [42]

Answer:

180

Step-by-step explanation:

formulae=H×L×W so....5×9×4=180

4 0

3 years ago

H(x)=1/8x^3-x^2<br><br> Over which interval does h have a positive average rate of change?

lara31 [8.8K]

Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

Step-by-step explanation:

$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

$\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\$

6 0

3 years ago