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3 years ago

Solve for a, 7-a=3 A.10 B.4 C.2 D5

Mathematics

2 answers:

Tcecarenko [31]3 years ago

5 0

Answer:

Step-by-step explanation:

Vika [28.1K]3 years ago

5 0

7 - a + 3

subtract 7 from both sides.

-a = 3 - 7

add like terms (3 + - 7)

-a = -4

divide both sides by -1. (remember that in front of the a, a 1 is there.)

a = 4

answer: b. 4

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saul85 [17]

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HELP. It’s LOG form. Since were out of school I forgot how I did this and it’s due so it would be a life saver.

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Answer:

see below

Step-by-step explanation:

exponent to log: $a^b$ = c ---> logₐc = b

ie. question 6

log ₁₀(3x+1) = 2 -----> $10^{2} = 3x+1$

that will get you through questions 1 to 3, 5 to 6, and 8

in question 4, all you have to do is know that 2^2 = 4 and 2^3 = 8, by setting the bases equal, you can manipulate the exponents to get 2x+8 = 3x-3

for questions 7 and 9,

remember that:

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3 years ago

What is the solution to the equation below?

Alekssandra [29.7K]

Answer :

It is D because you have to subtract 4.73 from both sides in order to isolate the y by itself and get the answer which is 3.27

Step-by-step explanation:

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2 years ago

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Write the equation of the line that passes through (-2, 6) and (2, 14) in slope- intercept form.

creativ13 [48]

Answer:

Y= 2x+10

Step-by-step explanation:

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3 years ago

Find all solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically. (Enter your answer

Anuta_ua [19.1K]

Answer:

The solutions of the equation are 0 and 0.75.

Step-by-step explanation:

Given : Equation $16x^4 - 24x^3 +9x^2 =0$

To find : All solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically ?

Solution :

Equation $16x^4 - 24x^3 +9x^2 =0$

$x^2(16x^2-24x+9)=0$

Either $x^2=0$ or $16x^2-24x+9=0$

When $x^2=0$

$x=0$

When $16x^2-24x+9=0$

Solve by quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-24)\pm\sqrt{(-24)^2-4(16)(9)}}{2(16)}$

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For verification,

In the graph where the curve cut x-axis is the solution of the equation.

Refer the attached figure below.

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