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serg [7]
3 years ago
10

Jada walks to school. The function D givers her distance from school, in meters, t minutes since she left home. Which equation t

ells us “Jada is 600 meters from school after 5 minutes”? A. D(600)=5 B. t(5)=600 C. t(600)=5 D. D(5)=600
Mathematics
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer:

t(5)=600.................

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A hotel Manager needed to seat 250 guests at a wedding party. Can he seat them in groups of 5 or 10?
svet-max [94.6K]

The manager can do both but if he does 10 he will have 25 groups but if he does 5 he will have 50 groups so the simple way is 10 so he will have 25 groups

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4 years ago
Coach Henderson had 16 gallons of paint to use an equal amount of paint each five days how many gallons do you use in each day
Veronika [31]

Answer:

16 ÷ 5 = 3.2 gallons

Step-by-step explanation:

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3 years ago
I need help finding v
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Check the picture below.

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4 years ago
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

3 0
3 years ago
Read 2 more answers
- 8/9x - 4/45x + 1/5x = - 70
lara31 [8.8K]

Answer:

x = 90

Step-by-step explanation:

- \frac{8}{9} x - \frac{4}{45} x + \frac{1}{5} x = - 70

Multiply through by 45 ( the LCM of 9, 45, 5 ) to clear the fractions

- 40x - 4x + 9x = - 3150 , that is

- 35x = - 3150 ( divide both sides by - 35 )

x = 90

8 0
3 years ago
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