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frozen [14]
3 years ago
5

Help!! Algebra help will give stars

Mathematics
1 answer:
Inessa [10]3 years ago
5 0

Answer:

I'll answer one question

Step-by-step explanation:

a. x >= -4. This reads, the domain is greater than and equal to negative four to infinity.

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Round 36.11 to the nearest tenth
Vilka [71]
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<em>5 or more round up, 4 or less round down
</em>As 1 is lower than 4, we round down.
36.11 therefore rounds down to 36.1
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Say that you own property with a market price of $74,000. The state tax assessors have given it an assessed value of 48% of that
BigorU [14]

Answer:

A

Step-by-step explanation:

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5 0
3 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
3 years ago
The point (-7,4) is reflected over the line x=-3. Then the resulting point is reflected over the line y=x. Where is the point lo
Ne4ueva [31]

Answer:

  (4, 1)

Step-by-step explanation:

Since the first reflection is over the vertical line x=-3, the y-coordinate remains the same. The x-coordinate of A' will make the point (-3, 4) on the line of reflection be the midpoint between A and A':

  (-3, 4) = (A +A')/2

  2(-3, 4) -A = A' = (-6-(-7), 8 -4) = (1, 4)

The reflection over the line y=x simply interchanges the two coordinate values:

  A'' = (4, 1)

3 0
3 years ago
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