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3 years ago

11

Evaluate: Percents Benita answered 85% of the test questions correctly. If she answered 9 questions incorrectly, how many questi

ons were on the test?
It is urgent.

1 answer:

madreJ [45]3 years ago

3 0

Answer:

60

Step-by-step explanation:

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80% of what number is 190

Degger [83]

80% of 190 = 152
THE ANSWER IS 152

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3 years ago

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Joe and Mark wash and wax cars on the weekend. On each car, they spend $1.50 on soap, $3.00 on wax, and $1.25 on wheel & tir

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Their total expenses are ...
.. $1.50 +3.00 +1.25 = $5.75

Their profit is the difference between their charge and their cost:
.. $25.00 -5.75 = $19.25

The correct choice is ...
B) $19.25

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3 years ago

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Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

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3 years ago

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4x + 1 over 3 y2

vampirchik [111]

Answer:

1/3 =0.33

Step-by-step explanation:

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<em>4</em><em>(</em><em>2</em><em>)</em><em>+</em><em>1</em><em> </em><em>/</em><em>3</em><em>(</em><em>3</em><em>)</em><em>^</em><em>2</em>

<em>=</em><em>8</em><em>+</em><em>1</em><em> </em><em>/</em><em>3</em><em>×</em><em>9</em>

<em>=</em><em>9</em><em>/</em><em>2</em><em>7</em>

<em>=</em><em>1</em><em>/</em><em>3</em>

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4 years ago

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1. Are these two lines parallel, perpendicular or neither?

kolezko [41]

Answer:

parallel

Step-by-step explanation:

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