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NikAS [45]
3 years ago
10

Which equation is the inverse of 5y+4 = (x+3)^2 +1/2?

Mathematics
2 answers:
maxonik [38]3 years ago
7 0

Answer:

Answer D

Explanation:

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To simply find the answer, you need to a: solve for y ; b: Find the square root of both sides ; c: solve the equation.

Hope this helps.

lora16 [44]3 years ago
7 0

Answer:

The answer for this question is D..

Step-by-step explanation:

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Two angles are supplementary. The larger angle is 15 more than 10 times the smaller angle. FInd the measure of each angle.
garik1379 [7]

\text{Larger angle} =x\\\\\text{Smaller angle} =y\\ \\\text{According to the condition,}\\\\x=10y+15\\\\\text{The two angles are supplementary, so their  sum is}~ 180^{\circ} \\\\x+y=180\\\\\implies 10y+15 +y = 180\\\\\implies 11y = 180-15\\\\\implies 11y=165\\\\\implies y = \dfrac{165}{11}\\\\\implies y = 15\\\\\text{So,} ~x  =180-15 = 165\\\\\text{Hence the larger angle is}~ 165^{\circ}~ \text{and the smaller angle is }15^{\circ}

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2 years ago
What is an equation of the line that is parallel to y=4x−10 and passes through (1, 13) ?
klasskru [66]
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y = 4x + c
sub in coorfinates (1,13)
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3 years ago
you are sharing 2 loaves of bread with 5 friends. you want each person to get a fair share how much bread will each person get
natali 33 [55]
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3 years ago
Read 2 more answers
Find the value of expression - 17x-19xpwer4 - 21x power 2,when x=3<br><br>​
Marrrta [24]

Answer:

-17x - 19x^4 - 21x^2 = -1779 when x = 3

Step-by-step explanation:

Given

-17x - 19x^4 - 21x^2

x = 3

Required

Solve

Substitute 3 for x in -17x - 19x^4 - 21x^2

-17x - 19x^4 - 21x^2 = -17*3 - 19*3^4 - 21*3^2

Evaluate all exponents

-17x - 19x^4 - 21x^2 = -17*3 - 19*81 - 21*9

-17x - 19x^4 - 21x^2 = -51 - 1539- 189

-17x - 19x^4 - 21x^2 = -1779

7 0
3 years ago
If anyone can check my work for me that would be greatly appreciated I'm having trouble.
krek1111 [17]
The answer is -13. Solution: = |-4b - 8| + |-1 - b^2| + 2b^3 = |-4(-2) - 8| + |-1 - -2^2| + 2(-2)^3 = |8-8| + |-1+4| + 2(-8) = |0| + |3| + (-16) = -13
8 0
3 years ago
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