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kiruha [24]
3 years ago
6

Can someone help please?!! Major Help Pic above

Mathematics
1 answer:
Mila [183]3 years ago
8 0
I believe it’s c. I hope that this is helpful!
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Are these triangles congruent? If so, why?
N76 [4]
<h3>Yes the triangles are congruent by ASA axiom or congruency .</h3>

Because one side of triangle is equal and two angles are equal . So option 1 is your answer .

5 0
3 years ago
Y=3x+7 solve this system of equation using any method
love history [14]

Answer:

x=2, y=13

Step-by-step explanation:

In order to get this by addition which is the easiest way for this you want to take both equations

y=3x+7 and y=-4x+21 and move x to the other side of the equation so it is like  4x+y=21 and -3x+y=7 like that now you have to get rid of one of the exponents and the easiest one to get rid of is y so I'm going to multiply -3x+y=7 by negative 1 so it becomes 3x-y=-7 that way the equations value has not changed since I multiplied both sides and now when I add

4x+y=21 and 3x-y=-7 It'll look like

4x+y=21

3x-y=-7          add those It will get rid of the y

then get 7x=14 simplify to get x=2 then you can take -3x+y=7 plug in x and get

-3(2)+y=7 multiply into -6+y=7 then simplify you get y=13

3 0
3 years ago
7 less than twice a number is 3 more than the number.
lorasvet [3.4K]

Answer:

7-2x=3+x

Step-by-step explanation:

where x is the number

3 0
3 years ago
Please help me solve this ;-;
Digiron [165]

9514 1404 393

Answer:

  C  log3(√((x -4)/x)

Step-by-step explanation:

The applicable rules of logarithms are ...

  log(a/b) = log(a) -log(b)

  log(a^n) = n·log(a)

The base is irrelevant, as long as all logs are to the same base.

__

  \dfrac{1}{2}(\log_3{(x-4)}-\log_3{(x)})=\dfrac{1}{2}\log_3(\dfrac{x-4}{x})=\boxed{\log_3\left(\sqrt{\dfrac{x-4}{x}}\right)}

6 0
3 years ago
Find an equation of the tangent line to f(x) = 5x2 − 2x + 10 at x = 7.
laiz [17]

Answer:

y=68x-235

Step-by-step explanation:

f(x) = 5x^2-2x + 10

Differentiate:

\implies f'(x)=10x-2

Substitute x=7 into f'(x):

\implies f'(7)=10(7)-2=68

Therefore, 68 is the gradient of the tangent line at x=7:

\implies y-y_1=68(x-x_1)

when x=7, y=5(7)^2-2(7)+10=241

\implies y-241=68(x-7)

\implies y=68x-235

4 0
2 years ago
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