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balandron [24]
3 years ago
5

When n basketball uniforms are purchased, the cost, C, of each uniform is given by the equation.

Mathematics
1 answer:
balandron [24]3 years ago
5 0

Answer:

D. Subtraction property of equality; n = 15

Step-by-step explanation:

Given equation that shows the cost of n basketballs uniforms,

C = 40n + 260

According to the question,

Total cost = $ 860,

⇒ 40n + 260 = 860

Using subtraction property of equality, subtract 260 on both sides,

⇒ 40n = 860 - 260

⇒ 40n = 600

Using division property of equality, divide both sides by 40.

⇒ n = \frac{600}{40} = 15.

Hence, 15 uniforms were purchased.

i.e. Option D is correct.

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nevsk [136]
3t+10=16
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3T=6
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3 years ago
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1.<br> Find the slope of the line.<br> Ay<br> a) -574<br> b) 5/4<br> c) -4/5<br> d) 4/5
maria [59]

Answer:

a) -5/4

Step-by-step explanation:

Slope is fundamentally rise/run; The points on the line are relative to each other by decreasing/increasing the y-value of +/-5 and decreasing/increasing the x-value by +/-4

The function is decreasing so the slope has to be negative.

The only possible answer choice is -5/4 or a.

3 0
2 years ago
What is the value 0.25 ? The 2 has a value of 2 _____________ or 0.2 The 5 has a value of 5 ______________ or 0.05
3241004551 [841]

Answer:

NUT?

Step-by-step explanation:

6 0
2 years ago
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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
Is this equation linear: 2x+5xy-3=0
Lina20 [59]

What is linear equation?

An equation between two variables that gives a straight line when plotted on a graph.

2x + 5xy - 3 = 0

x (2 + 5y) -3 = 0

x ( 2 + 5y ) = 3

2 + 5y = 3 / x

5y = (3/x) - 2

y = ( (3/x) - 2) / 5

According to the graph i don't think it's a linear equation.

3 0
3 years ago
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