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3 years ago

Carmen made $247 for 13 hours of work.

Mathematics

1 answer:

Scrat [10]3 years ago

4 0

Answer:

9 hours

Step-by-step explanation:

$247/13hours=$19 per hour

$171/$19=9 hours

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3 years ago

Read 2 more answers

PLEASE HELP ME IM STRUGGLING!!!

Kitty [74]

Answer:

The required answer is $c=7\sqrt{3}$

Therefore the number in green box should be 7.

Step-by-step explanation:

Given:

AB = 7√2

AD = a , BD = b , DC = c , AC = d

∠B = 45°, ∠C = 30°

To Find:

c = ?

Solution:

In Right Angle Triangle ABD Sine identity we have

$\sin B = \dfrac{\textrm{side opposite to angle B}}{Hypotenuse}\\$

Substituting the values we get

$\sin 45 = \dfrac{AD}{AB}= \dfrac{a}{7\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}= \dfrac{a}{7\sqrt{2}}\\\\\therefore a=7$

Now in Triangle ADC Tangent identity we have

$\tan C = \dfrac{\textrm{side opposite to angle C}}{\textrm{side adjacent to angle C}}$

Substituting the values we get

$\tan 30 = \dfrac{AD}{DC}= \dfrac{a}{c}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{7}{c}\\\\\therefore c=7\sqrt{3}$

The required answer is $c=7\sqrt{3}$

8 0

3 years ago

Find the polynomial f(x) of degree 3 with real coefficients that has a y-intercept of 60 and zeros 3 and 1+3i.

Sauron [17]

$\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0$

[ correction added, Thanks to @stef68 ]

$\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)$

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3 years ago