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3 years ago

The GCF of 36 and 63 is A) 3 B) 9 C) 7 D)18

Mathematics

2 answers:

kifflom [539]3 years ago

7 0

The correct answer is B)9

crimeas [40]3 years ago

5 0

B cause if you do 9 x 4 you’ll get 36 and if you times 9 X 7 you’ll get 63

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A grocery store sells a bag 7 oranges for $2.66. if alexandra spent $3.42 on oranges how many did she buy

Murljashka [212]

Answer:

9 oranges

Step-by-step explanation:

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3 years ago

Read 2 more answers

How in the world do I answer this problem??

galben [10]

Answer:

Step-by-step explanation:

ST is y and SU is 2y-4.

Note that SU = ST + TU = y + 18

So we have 2y - 4 = y + 18

2y - 4 = y + 18

2y = y + 22

y = 22

So ST is 22.

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3 years ago

Which set of equations has (1, 2) as its solution?

FromTheMoon [43]

A and D... i think but i would rather go with A and C

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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