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Radda [10]
3 years ago
10

What is the equation of the line that passes through the point (-4,-3) and has a slope of 5?

Mathematics
1 answer:
Travka [436]3 years ago
5 0

Answer:

Y= 5x + 17

Step-by-step explanation:

Y-y1= m(x-x1)

Y- -3= 5(x- -4)

Y+ 3= 5x +20

Y= 5x + 17

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What is the volume of a hemisphere with a radius of 2.3m round to the nearest tenth of a cubic meter
kodGreya [7K]

Answer: 25.5 m^{3}

Step-by-step explanation:

If the volume of a sphere is

V=\frac{4}{3} \pi r^{3}

Where r=2.3 m is the radius

The volume of a hemisphere is half the volume of the total sphere:

V_{hemisphere}=\frac{V}{2}=\frac{\frac{4}{3} \pi r^{3}}{2}

V_{hemisphere}=\frac{2}{3} \pi r^{3}

Solving this equation:

V_{hemisphere}=\frac{2}{3} \pi (2.3 m)^{3}

Finally:

V_{hemisphere}=25.48m^{3} \approx 25.5 m^{3}

5 0
3 years ago
13 subtracted from a number w is 15.
yarga [219]
W-13=15|+13
w=15+13
x=28

<em>Answer: w=28</em>
7 0
3 years ago
Read 2 more answers
Analyze the diagram below and complete the instructions that follow.
juin [17]

Answer:

Hello

Step-by-step explanation:

the Pythagoras Theorem proves this

7 0
3 years ago
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Can anyone help me with #14?
liraira [26]
Number of students on a sports team
8 0
3 years ago
Rationalize the denominator.
Maslowich
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166412

——————————
 
     \mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}


Multiply and divide by the conjugate of the denominator, which is \mathsf{\sqrt{a}+2\sqrt{y}:}

     \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}

The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:
 
     \mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}} 

     \mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

8 0
3 years ago
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