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qaws [65]
3 years ago
6

Tell whether the angles are adjacent or vertical. Then find the value of x.

Mathematics
2 answers:
goldenfox [79]3 years ago
8 0
The angles are verticals and the value of x is 128°.
SVEN [57.7K]3 years ago
8 0
Answer:

The angles are vertical
X= 128

Hope that helps :)
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To rent a certain meeting room, a college charges a reservation fee of $39 and an additional fee of $5.80 per hour. The film clu
garri49 [273]

Answer:

5 hours

Step-by-step explanation:

$39 + $5.8*t < $73.80

To solve it, collect the constant terms on the right side:

5.8*t < $73.80 - $39 = $34.08

t < $34.08 divided by 5.8 = 5 hours.

Idk if this is correct, I had a question like this on a test and this is what I did. So plz don't be mad if it's wrong.

4 0
3 years ago
Abby packs plastic cubes in a box. Each cube is 1 square centimeter.
ira [324]

Answer:

2640 cubes

Step-by-step explanation:

20 cm x 12 cm x 11 cm = 2640 cm^3

5 0
3 years ago
Which is equal to 4 hectometers?
SCORPION-xisa [38]

Answer:

B.)

Step-by-step explanation:

baby :p

6 0
3 years ago
Read 2 more answers
CAN SOMEONE HELP WITH THIS ASAP
Mazyrski [523]

Answer:

When x=1/2, y=28/3

when x =0, y=9

Equation is y=2/3 x + 9

Step-by-step explanation:

First lets convert them to fractions to make it easier. Also put them in order.

X                     Y

0

1/2  

3/4                  19/2

1                      29/3

6/5                 49/5

You can find the slope since we know 2 points.

(3/4, 19/2) and (6/5, 49/5)

m=3/10 / 9/20

m=2/3

y=2/3 x + b

Lets plug in x and y points to find the y-int

19/2=2/3(3/4) + b

b=9

Therefore y=2/3 x + 9

When x=1/2, y=28/3

when x =0, y=9

8 0
3 years ago
Given the position of the particle, what the position(s) of the particle when it’s at rest
choli [55]

The position function of a particle is given by:

X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t

The velocity function is the derivative of the position:

\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}

The particle will be at rest when the velocity is 0, thus we solve the equation:

2t^2-9t-18=0

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Substituting:

\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}

We have two possible answers:

\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

undefined

6 0
2 years ago
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