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FinnZ [79.3K]
2 years ago
7

What is the slope of line x -3,-2,-1,0 y 6,4,20? PLEASE HELP ASAP

Mathematics
1 answer:
alekssr [168]2 years ago
7 0
Answer: y=-2x
Solution:

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Glven: 2x+3y= 6.
ruslelena [56]

Answer:

The answer is B 0,2

Step-by-step explanation:

I hope that correct

7 0
3 years ago
If a1=4 and an =2an-1 – 1 then find the value of a4.
musickatia [10]
1 = 4

a2 = -2(a1) - 1
a2 = -2(4) - 1
a2 = -9

a3 = -2(a2) - 1
a3 = -2(9) - 1
a3 = -19

Can you do a4 and a5 and answer?
7 0
2 years ago
Read the following statement. It is cloudy outside. What is the first step of indirectly proving this statement?
EleoNora [17]
Assume if it is cloudy, then you are outside.
5 0
2 years ago
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The length of a photograph is 3cm less than twice the width. The area is 54cm². Find the dimentions of the photograph.
AnnyKZ [126]
L=2w-3 

w(2w-3)=54

2w²- 3w-54=0

A= 2 , B= -3, C= -54

sub values into quadratic formula
 i attached picture of formula

Final answer:  you will be given 2 answers 6 and -4.5. -4.5 is rejected since a negative value cannot represent length because that just doesn't make sense.

w=6

Check:
L=2w-3
=2(6)-3
=12-3
=9
  9*6=54

Therefore the dimensions of the photograph are 6 cm by 9 cm



3 0
3 years ago
Read 2 more answers
Find the volume and area for the objects shown and answer Question
klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of ​​the given solids.

\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2

\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\

\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}

\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}

SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}

6 0
2 years ago
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