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3 years ago

8

the jacobys went on a 600- mile trip. On the first day they drove 52/3 hours and on the second day they drove 4 3/5 hours how ma

ny hours did they drive during the first two days

2 answers:

Alexxx [7]3 years ago

8 0

The answer is:10 hrs and 48 min. Hope this helps! (:

svetlana [45]3 years ago

7 0

I believe the answer would be 10 hours and 48 minutes.

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99 X 99 X 99 / 9 - 99 /99 PLEASE EXPLAIN ;)

charle [14.2K]

Answer:

107810

Step-by-step explanation:

You can use the order of operations! So we start with 99^3/9. That is 99^2*11. So now let’s do 99/99, it’s 1. So 99^2*11 is 107811, so the answer is 107810.

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3 years ago

Read 2 more answers

Please Help I Don't Undertand.

frozen [14]

If triangles AMN and ABC are similar, then

AM/AB = AN/AC

or

AM/(AM + MB) = AN/(AN + NC)

Check if this is true:

AM/AB = 21/(21 + 9) = 21/30 = 7/10

AN/AC = 14/(14 + 6) = 14/20 = 7/10

The angle at vertex A is common to both of the triangles.

Then by the side-angle-side (SAS) similarity theorem, the triangles are indeed similar.

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1 year ago

How many solutions does the equation 3x − 7 = 4 + 6 + 4x have? Two Zero Infinitely many One

sleet_krkn [62]

Answer:

one solution

Step-by-step explanation:

3x − 7 = 4 + 6 + 4x

Combine like terms

3x-7=4x+10

Subtract 3x from each side

-7=x+10

Subtract 10 from each side

-17 =x

There is one solution

4 0

3 years ago

How to calculate the rise and run of a slope (4,-1) and (-2,3)

astra-53 [7]

$m = \frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }$
$m = \frac{3 - (-1)}{-2 - 4}$
$m = \frac{4}{2}$
$m = 2$

8 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2

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1 year ago