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lubasha [3.4K]
3 years ago
15

I WILL MARK BRINLIEST

Mathematics
1 answer:
lisov135 [29]3 years ago
7 0

Answer:

f

bicycle 24 x2= 48

tricycle 11 x3= 33

48+33=81

Step-by-step explanation:

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Need help asap!! will give brainliest!!
frozen [14]

Answer:

X and Y = 4

Well, it's a right triangle. The red box means 90 degree angle. So I would say that all the sides are equal.

8 0
2 years ago
Read 2 more answers
Please help me..........<br><br>only 32,and 33​
kupik [55]

Answer:

Step-by-step explanation:

32

Step 1 : Draw PQ = 6cm

Step 2 : Construct 60 degree at Q. [Given PQR = 60=> Q = 60 ]

Step 3 : Take 5 cm on compass and mark 5cm on the 60 degree

            line constructed at Q.

Step 4 : The arc of 5cm and 60 degree line  meets is the point R.

             [Given QR = 5cm ]

Step 5 : Given PQ || SR so construct a 120° at R ,

             because ∠PQR and ∠QRS are supplementary angle.

            [ ∠PQR + ∠QRS = 60 + 120 = 180° ]

Step 6 : At P , take 6.5 cm on compass and mark on the line drawn in

             Step 5, to get S. [ Given PS = 6.5 ]

           

33

Given adjacent sides : 4.8 and 4.2

And it is a rectangle.

Step 1 : construct AB = 4.8 cm

Step 2 :At A and B construct 90 degree

Step 3: On the compass take 4.2cm and mark on the 90degree lines

          from A and B.

Step 4 : Mark that as C and D respectively.

Step 5 : Join CD

8 0
3 years ago
Please need help, Will give brainiest!!!!
timurjin [86]
I'd say its D but im not quite sure, its what makes most sense to me.
6 0
3 years ago
to rent a building for a school dance ava paid $120 plus $2.50 for each student who attended. If she paid a total cost of $325 h
deff fn [24]

Answer:

82 students

Step-by-step explanation:

8 0
3 years ago
A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
qaws [65]

The projectile's horizontal and vertical positions at time t are given by

x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t

y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2

where g=9.8\dfrac{\rm m}{\mathrm s^2}. Solve y=0 for the time t it takes for the projectile to reach the ground:

30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s

In this time, the projectile will have traveled horizontally a distance of

x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m

The projectile's horizontal and vertical velocities are given by

v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ

v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about \sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}.

7 0
3 years ago
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