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Vedmedyk [2.9K]

2 years ago

14

Mikayla caught a value pack of crackers for $6.72. The value pack had 24 individually wrapped crackers packages. Solve 24x = $6.

72 to find the cost per package in dollars.

1 answer:

Licemer1 [7]2 years ago

7 0

Answer:

0.28$

Step-by-step explanation:

divide

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What is the answer to: 3 x 4 + 15- 19 x2037? help

Ierofanga [76]

Answer: -38,686

Step-by-step explanation:

3 x 4= 12

-19 x 2037= -38,703

12 + 15= 17

17 -38,703= -38,686

6 0

3 years ago

crater Lake in Oregon in shaped like a circle with a diameter of about 5.5 miles. What is the area of the surface of crater lake

N76 [4]

Answer:

Given the statement: Crater Lake in Oregon in shaped like a circle with a diameter of about 5.5 miles.

Area of circle is given by:

$A = \pi \frac{d^2}{4}$

where

A represents the area of a circle

d represents the diameter of a circle.

then;

It is given that Surface of Crater lake is shaped like a circle.

Substitute the given values, we have;

Area of a circle = $\pi \frac{(5.5)^2}{4}$

= $3.14 \times \frac{30.25}{4} =3.14 \times 7.5625 = 23.74625$ square miles

therefore, the area of a surface of a crater lake is, 23.74625 square miles.

3 0

2 years ago

BC = Round your answer to the nearest hundredth

stich3 [128]

What am I’m supposed to be looking at

3 0

3 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,

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

Hence, the midpoint of the segment is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

6 0

2 years ago

Please help! I attached the question below.

kompoz [17]

Answer:

$\frac{2(c+2)}{c(c-2)}$

Step-by-step explanation:

$\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }$

Identity used:

$a^{2}-b^{2}=(a-b)(a+b)$

$\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }$

Now let us divide the modified expressions:

$\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})}$ ÷ $\frac{(c-2)^2}{2(6c^{3}+15c^{2}) }$

we get:

$\frac{2(c+2)}{c(c-2)}$

5 0

3 years ago