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tatuchka [14]
3 years ago
10

PLEASE HELP ASAP!! I NEED TO GET MY MATH GRADE UP! I WILL MARK BRAINLIEST!

Mathematics
2 answers:
rodikova [14]3 years ago
8 0

Answer:

The y intercept would be 1,0

Step-by-step explanation:

Hope it helped

Arturiano [62]3 years ago
7 0

Answer:

10.6

Step-by-step explanation:

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What is the volume of the cone? (round to the nearest tenth)
Ipatiy [6.2K]

Answer:

a. 402.1 cubic units

Step-by-step explanation:

volume of cone = 1/3 × πr²h

we have height already which is 6

for radius we use pythagoras theorem, and find out the radius is √10²-6² = 8

putting these values in the formula, we get the volume.

3 0
3 years ago
Select the correct answer from each drop-down menu.
eimsori [14]

Answer:

(L+1)/2 = W

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Which equation represents a function that is nonlinear?
Alenkasestr [34]

Don't hate on me if i'm wrong please...

Answer: y=x^2-25 or C

7 0
3 years ago
Find tan2θ if θ terminates in Quadrant IV and cosθ =3/5 .
Liula [17]

Answer:

The answer to your question is  \frac{-8}{3}

Step-by-step explanation:

Data

cos Ф = 3/5

tan2Ф = ?

Quadrant IV

Process

1.- Remember that tanФ = \frac{Opposite side}{Adjacent side}  and cos Ф = \frac{adjacent side}{hypotenuse}

then,    adjacent side = 3 and hypotenuse = 5

2.- Calculate opposite side

             O² = h² - a²

             O² = 5² - 3²

             O² = 25 - 9

              O² = 16

              O = 4

3.- Calculate the tan 2Ф

            tan 2Ф = 2(\frac{-4}{3})       The opposite side is negative in the fourth

                                           quadrangle

            tan 2Ф = \frac{-8}{3}

6 0
4 years ago
1. A family is building a circular fountain in the backyard. The yard is rectangular and measures 10x by 15x and the fountain is
sp2606 [1]
1. The remaing area of the yard will be found as follows:
Area of the yard

A1=length*width=10x*15x=150x^2 yd^2

area of the fountain
A2=πr²
A2=π(4x)²
=50.266x^2

The remaining area will be:
A=150x^2-50.266x^2
A=99.735x^2 yd^2


2] Area left for bleachers, restrooms and other parts of the stadium will be as follows:
Area of the lot is:
A1=length*width
A1=8x×12x= 96x^2 

Area of the field
A2=length×width
A2=3x×6x=18x^2

Hence the remaining area will be:
A1-A2
=96x^2-18x^2
=78x^2 sq. units


7 0
3 years ago
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