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Lena [83]
2 years ago
10

Anyone know this??? Pls help now.

Mathematics
1 answer:
Brut [27]2 years ago
4 0

<em>x,y = (3,-7)</em>

<em>x-5, 7+4</em>

3-5 = -2

-7+4=-3

<em>the answer is (-2,-3)</em>

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_ 9.5:12.6 That is the answer, btw the 6 is repeating
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Solve x3 = 1/8. Please help Will mark brainliest
I am Lyosha [343]

Answer:

here's the steps

Step-by-step explanation:

first, you need to get x by itself.In order to do that you need to do the opposite to the other side. ex: x5= 1/10; what you would do is divide 1/10 by 5 to get x = 0.5 , 5/10

5 0
3 years ago
Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k​
valkas [14]

Answer:

\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k​, the vector projection of B unto a is expressed as proj_ab = \dfrac{b.a}{||a||^2} * a

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k  =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||²  = 130

proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\

<em>Hence the projection of b unto a is expressed as </em>\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\<em></em>

7 0
3 years ago
PLEASE HELP
Aneli [31]

Answer:

Terrell drove 200 miles one way

Step-by-step explanation:

Since his trip concluded at 9 hours, let's say that x = time traveled at 40 mph

and 9 - x = time traveled at 50 mph

D = (rate) x (time)

<em>40x = 50(9 - x)</em> -----Use Distributive property on the right side of the equation

<em>40x = 450 - 50x</em><em> </em>-----Add 50x on both sides

<em>90x = 450</em> ------Divide by 90 on both sides

We end up with<em>  x = 5  </em>which is the value we need to determine the distance of our first trip, calculated by

<em>D = 40x</em>, or <em>40(5) = 200 mi</em>

6 0
3 years ago
Numerical value of cosh(ln5)
strojnjashka [21]
\cosh x=\dfrac{e^x+e^{-x}}2

So

\cosh(\ln 5)=\dfrac{e^{\ln5}+e^{-\ln5}}2=\dfrac{5+\frac15}2=\dfrac{13}5
6 0
3 years ago
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