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podryga [215]
2 years ago
10

Solve 4X squared minus X -5 equals zero

Mathematics
1 answer:
Anettt [7]2 years ago
7 0

your answer should be 1/3 if I did my math right

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They are similar and the ratio is still the same. The artwork's shape didn't change we just resized it by multiplyting its length and width by 2.

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Help please I’ve been looking online for help but couldn’t find any.
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A college basketball team offers season passes for $85, but you pay $4.45 for a program
oksano4ka [1.4K]

For 9 of games is the cost of these plans the same.

<h3>What are basic linear equations?</h3>

Moving the variables to one side of the equation and the numeric portion to the other allows us to solve a linear equation with one variable. As an illustration, the equation x - 1 = 5 - 2x can be resolved by transferring the numerical components to the right side of the equation while leaving the variables on the left. Therefore, we receive x + 2x = 5 + 1. Thus, 3x Equals 6. As a result, x = 2. The formula for a two-variable linear equation is Ax + By + C = 0, where A and B are the coefficients, C is a constant term, and x and y are the two independent variables, each with a degree of 1. A linear equation involving two variables, for illustration, is 7x + 9y + 4 = 0.

Consequently, you would multiply $9.45 and $4.45 by the same number (I selected 9 because there are only one or two possible answers), then add $85 to the result for the $4.45 amount.

9.45 x 9

= 85.05

4.45 x 9

= 40.05 +85

= 125.05

To know more about linear equation ,visit:

brainly.com/question/11897796

#SPJ13

4 0
1 year ago
Determine the value of variables a, b, and c that make each equation true.
dybincka [34]

Corrected Question

Determine the values of a, b and c that make each equation true.

(x^a)^6=\dfrac{1}{x^{30}} \\\\(x^{-7})^{-4}=x^b\\\\(x^{-2})^c=x^{22}

Answer:

a=-5, b=28 and c=-11

Step-by-step explanation:

To solve for a,b and c, we apply the following laws of indices

\dfrac{1}{x^y}=x^{-y} \\\\(x^m)^n=x^{m X n}\\\\$If x^m=x^n,$ then m=n

Therefore

(x^a)^6=\dfrac{1}{x^{30}}\\x^{a*6}=x^{-30}\\6a=-30\\a=-5

To solve for b

(x^{-7})^{-4}=x^b\\x^{-7*-4}=x^b\\x^{28}=x^b\\b=28

To solve for c

(x^{-2})^c=x^{22}\\x^{-2*c}=x^{22}\\-2c=22\\c=-11

4 0
3 years ago
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