Can someone help me with the middle question?

Mathematics

1 answer:

kotegsom [21]3 years ago

7 0

You should do 1.0 divided by 7 and the answer would be 7

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9/x+7 = 4<br> i need the answer for my homework, please help x

Pachacha [2.7K]

Your answer is Answer:
x = -3

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2 years ago

Which of the following sets of numbers could not represent the three sides of a right<br> triangle?

nirvana33 [79]

The

Answer: number 3

Explain:

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2 years ago

A farmer sells 8.7 kilograms of pears and apples at the farmer's market. 2 5 of this weight is pears, and the rest is apples. Ho

Ivenika [448]

Answer: 5.22kg

Step-by-step explanation:

From the question, a farmer sells 8.7 kilograms of pears and apples at the market. Of the pears and apples sold, 2/5 of this weight is pears, and the rest is apples. To calculate the kilograms of apple sold goes thus:

Total kilograms sold = 8.7kg

Fraction of pears weight = 2/5

Pears kilograms sold = 2/5 × 8.7

= 0.4 × 8.7

= 3.48kg

Since the kilograms of pears sold is , 3.48kg, to get the kilograms of apple sold, we subtract 3.48kg from 8.7kg. This will be:

= 8.7kg - 3.48kg

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2 years ago

Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)

JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by:

$\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9$

Plugging in the values in formula, we get,

$\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}$