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3 years ago

What’s the answer to the problem?

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

6 0

Answer:

Step-by-step explanation:

if you do the checker method then you would find that

they add up to make a line which equals 180

so... the equation is (8x-7)+(x+16)=180

do the math and x=19

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?/5 =8/10 type the missing number that makes this fraction equal

SOVA2 [1]

<h2>- - - - - - - - - - - - - - - - - - </h2><h3> $\frac{?}{5} = \frac{8}{10}$ </h3><h3>Let's take the missing number as x.</h3>

<h3> $\frac{x}{5} = \frac{8}{10}$ </h3><h3>Means = Extremes </h3><h3> $x \times 10 = 5 \times 8$ </h3><h3> $x = \frac{5 \times 8}{10}$ </h3><h2>___</h2><h3>(Cancel by 2 table)</h3><h3> $x = \frac{5 \times4 }{5}$ </h3><h2>___</h2><h3>(Cancel by 5 table, So that we cancel 5 in the denominator and numerator)</h3><h3> $x = 4$ </h3><h2>- - - - - - - - - - - - - - - - - -</h2>

<h2>Therefore, The missing number is 4. </h2>

5 0

2 years ago

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Please please help me

IgorLugansk [536]

1. 8.75
2. 2
3.5+4.5= 9.5

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3 years ago

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What is the name of the tide when the sun, moon, and earth are all aligned, causing the greatest pull for highest and lowest tid

Naily [24]

Answer:

Whenever the earth, moon and sun are aligned, the gravitational pull of the moon is augmented by the gravitational pull of the sun, causing larger than average tidal ranges — meaning especially high high tides and very low low tides. These extreme tides are called “spring tides.”

Step-by-step explanation:

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3 years ago

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Find the missing side. Round to the nearest tenth.

AnnZ [28]

Answer:

B. 29.5

Step-by-step explanation:

Cos=A/H

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3 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

1 year ago