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Karo-lina-s [1.5K]
3 years ago
7

Please I need help !!!!

Mathematics
1 answer:
frez [133]3 years ago
3 0

Answer:

The first one is c the second is b

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I need help with these 2 problems I cant figure out.
Alla [95]

I started by labeling the right angle (Angle C) 90º. Next, I wrote down everything in one equation.

2x + 90 + 3x - 20 = 180º (180 degrees in a triangle)

Next, I add 20 on both sides.

2x + 90 + 3x = 200º

I combine like terms (2x and 3x)

5x + 90 = 200º

I subtract 90 from both sides.

5x = 110º

Divide 110 by 5 to get x.

x = 22

======

For problem two, I label all the angles I know.

49º + 80º + r = 180º

I add 80 and 49.

129º + r = 180º

I subtract 180 and 129 and get 51º, which is your angle for R.

For angle X, you know that angle R plus angle X equals half of a circle, which is 180º

We know from before that 129º is 180º without R, so X is 129º

I hope this helps! Let me know if I'm wrong!

5 0
3 years ago
Wendy purchased 9 pizzas for x dollars. She spent a total of $85.50. Find the value of x
disa [49]
X=9.5 all you do is divide the total by the other factor you have
5 0
3 years ago
Read 2 more answers
At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is
garik1379 [7]
Check the picture below.

now, keep in mind that ship B is going at 20kph, thus from noon to 4pm, is 4 hours, so it has travelled by then 20 * 4 or 80 kilometers, thus b = 80.

whilst the ship B is moving north, the distance "a" is not really changing, and thus is a constant, that matters because the derivative of a constant is 0.

\bf c^2=a^2+b^2\implies \stackrel{chain~rule}{2c\cfrac{dc}{dt}}=0+2b\cfrac{db}{dt}\implies \cfrac{dc}{dt}=\cfrac{b\frac{db}{dt}}{c}
\\\\\\
\begin{cases}
\frac{db}{dt}=20\\
c=10\sqrt{353}\\
b=80
\end{cases}\implies \cfrac{dc}{dt}=\cfrac{80\cdot 20}{10\sqrt{353}}\implies \cfrac{dc}{dt}=\cfrac{160}{\sqrt{353}}
\\\\\\
\textit{and rationalizing the denominator}\implies \cfrac{160\sqrt{353}}{353}

8 0
3 years ago
Please Help me!!!
drek231 [11]

Step-by-step explanation:

The displacement of a particle d (in km) as a function of time t (in hours) is given by :

d=2t^3+5t^2-3

Displacement at t = 4 hours,

d(4)=2(4)^3+5(4)^2-3=205\ km

Velocity of particle is given by :

v=\dfrac{dd}{dt}\\\\v=\dfrac{d(2t^3+5t^2-3)}{dt}\\\\v=6t^2+10t

Velocity at t = 4 hours,

v=6(4)^2+10(4)=136\ km/h

Acceleration of the particle is given by :

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6t^2+10t)}{dt}\\\\a=12t+10

At t = 4 hours,

a=12(4)+10=58\ km/h^2

Therefore, the displacement, velocity and acceleration at t = 4 hours is 205 km, 136 km/h and 58 km/h² respectively.

7 0
3 years ago
D+(-9)=-5<br><br> plzzz help me
seropon [69]

Answer:\

the answer is D=4


4 0
4 years ago
Read 2 more answers
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