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Rina8888 [55]
3 years ago
11

How do I find the value of z?

Mathematics
1 answer:
Shalnov [3]3 years ago
3 0
Hello!

The answer is z = 16..

Solve it like this..

#1) Regroup terms

-3 = -7+z/4 ---> -3 = z/4 - 7

#2) Add 7 to both sides.

-3 = z/4 - 7 ---> -3 + 7 = z/4

#3) Do the addition.

-3 + 7 = z/4 ---> 4 = z/4

#4) Multiply both sides by 4.

4 = z/4 ---> 4 * 4 = z.

#5) Do the multiplication.

4 * 4 = z ---> 16 = z

Hope this helps! ☺♥
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Please help<br> Find the area in square units of ABC△ plotted below.
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Answer:

Use the distance formula on both points AC and AB.

<em>Distance formula is this</em><em>:</em>

<em>\begin{gathered}d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\\\d=\sqrt{(1--5)^2+(8--7)^2} \\\\d=\sqrt{(6)^2+(15)^2} \\\\d=\sqrt{36+225} \\\\d=\sqrt{261} \\\\\end{gathered}d=(x2−x1)2+(y2−y1)2d=(1−−5)2+(8−−7)2d=(6)2+(15)2d=36+225d=261</em>

Distance for AC  is  16.16

Now do the same with the numbers for AB and get the distance of 5.39

2. To get the area, use the formula 1/2 x base x height

AB is the base and AC is the height.

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6 0
2 years ago
Could someone do all the problems and/or check my work?
MArishka [77]
11. You’ve done it correctly

12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2

13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution

14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath

15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4

Hope this helped :)
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B is correct, and D is correct.

that is the answer to that question. Ive already done this question before and I got it right everytime

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