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fenix001 [56]
3 years ago
11

Which point is on the interior of ∠CAB?

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
6 0

The interior of an angle would be anything inside the lines of the angle in the direction of the arrow.

In this case the arrows are pointing downward, so Point E would be on the interior.

You might be interested in
Can you help me answer this question
r-ruslan [8.4K]
Division of a fraction is the equivalent of multiplying by its reciprocal
ex.
\frac{\frac{A}{B} }{\frac{C}{D}}=\frac{A}{B} *\frac{D}{C}
i suggest you try to remember this concept

in terms of your question

\frac{\frac{4c-12}{4c+8}}{\frac{c-3}{c^2-4}}= \frac{4c-12}{4c+8} *\frac{c^2 -4}{c-3}

from this point, just multiple the numerators together and denominators together, then simplify if necessary

fyi- difference of squares
x^2-b^2 = (x-b)(x+b)
relating that to
c^2-4 = c^2 -2^2 = (c-2)(c+2)

also a side note, you might want to factor out the 4 first in the top fraction


4 0
4 years ago
A cubical water tank can contain 1000/125 cubic meters of water. Find the length of a side of the water tank.
Annette [7]

Given:

Volume of cubical tank = \dfrac{1000}{125} cubic meters.

To find:

The length of a side of the water tank.

Solution:

The volume of a cubical tank is:

V=a^3

Where, a is the side length.

It is given that the volume of cubical tank is \dfrac{1000}{125} cubic meters. So,

a^3=\dfrac{1000}{125}

a^3=\dfrac{(10)^3}{5^3}

Taking cube root on both sides, we get

a=\dfrac{10}{5}

a=2

Therefore, the length of a side of the water tank is 2 meters.

3 0
3 years ago
Last year the 7th grade class had 350 students. This year the number decreased 36%. How many students are in this year's 7th gra
Hoochie [10]

Answer:

224

Step-by-step explanation:

350-36%=224. easy as that.

6 0
3 years ago
Write the equaition in slope-intercept form of the line through (6,9) and (-12,-3)
maxonik [38]
The answer to this question is y = 2/3x + 5
8 0
3 years ago
PLEASE HELP VERY URGENT!!!!!!!
Ierofanga [76]

Answer:

The correct option is;

C. (1.6, 1.3)

Step-by-step explanation:

Given that at x = 1.5 the y-values of both equations are y = 1.5 and y = 1 respectively

The x-value > The y-value

The difference in the y-values = 1.5 - 1 = 0.5

At x = 1.6 the y-values of both equations are y = 1.2 and y = 1.4 respectively

The x-value > The y-value

The difference in the y-values = 1.2 - 1.4 = -0.2

At x = 1.7 the y-values of both equations are y = 0.9 and y = 1.8 respectively

The x-value > The first y-value and the x-value < the second y-value

The difference in the y-values = 0.9 - 1.8 = 0.9

Therefore, the approximate y-value can be found by taking the average of both y-values when x = 1.6 where the difference in the y-values is least as follows;

Average y-value at x = 1.6 = (1.2 + 1.4)/2 = 1.3

Therefore, the best approximation of the exact solution is (1.6, 1.3)

By calculation, we have;

-3·x + 6 = 4·x - 5

∴ 7·x = 11

x = 11/7 ≈ 1.57

y = 4 × 11/7 - 5 ≈ 1.29

The solution is (1.57, 1.29)

8 0
3 years ago
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