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Nataly_w [17]
4 years ago
12

Write the product as a mixed number 5x1 7/8​

Mathematics
2 answers:
Naddik [55]4 years ago
4 0

Answer:

10 5/8

Step-by-step explanation:

Mars2501 [29]4 years ago
4 0
Step one change to decimal: 1 7/8=1.88
Step two multiply:1.88*5=9.4
Step three change back to fraction:9.4=9 2/5
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What is the value of f(16) - f(0) when f(x) = 4x - 8?
levacccp [35]
That would be 4(16) - 8 - (4(0) - 8) = 64 - 9 - 0 + 8 = 63 Answer
5 0
3 years ago
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If the current temperature is -11°F, what change in temperature would result in a final temperature of -27°F?
kirill [66]
-7 F I think is the answers
4 0
3 years ago
Do the ratios 5:3 and 13:8 form a proportion?
shepuryov [24]

Answer:

  No

Step-by-step explanation:

Each ratio is irreducible, and they are different. Hence they are not the same value, so cannot be a proportion. A proportion is two equal ratios.

  5 : 3 = 40 : 24

  13 : 8 = 39 : 24 . . . . not the same as 40 : 24

_____

<em>Examples of proportions</em>

  5/3 = 40/24 . . . is a proportion

  13/8 = 39/24 . . . is a proportion

4 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
Anyone please help me pleaseeee
lora16 [44]

R = { (x,y): 3x-y=0 }

The condition is 3x=y so that's not going to be any of these things.

R is reflexive if (x,x)∈R for all x.   Let's check.

3x - y = 3x - x = 2x ≠ 0  necessarily.   NOT REFLEXIVE

R is symmetric if (x,y)∈R → (y,x)∈R.   Let's check.

(x,y)∈R  so

3x-y = 0

y  = 3x

Is  (y,x)∈R.  That would be true if 3y-x=0

3y - x = 3(3x) - x = 8x ≠ 0 necessarily NOT SYMMETRIC

R is transitive if (x,y)∈R and (y,z)∈R → (x,z)∈R.   Let's check.

3x-y = 0  so y=3x

3y-z = 0 so z=3y = 9x

3x - z = 3x - 9x = -6x ≠ 0 necessarily   NOT TRANSITIVE

6 0
3 years ago
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