All categories

3 years ago

Help plz asap find the area

Mathematics

2 answers:

quester [9]3 years ago

7 0

Answer:36 maybe

Step-by-step explanation:

4×3=12

12÷2=6

6×5=30

30+6=36

miv72 [106K]3 years ago

7 0

Area of the given figure is combined area of a rectangle and a triangle,

so, Area of figure =

area of rectangle + area of triangle

now,

Area of rectangle =

length × width

6 × 5

30 cm²

Area of triangle =

1/2 × base × height

1/2 × 4 × 3

6 cm²

Area of whole figure =

30 cm² + 6 cm²

36 cm²

You might be interested in

Please help, I need Answers quickly.

Deffense [45]

Add 6.5 to both sides

D=-0.7

A

8 0

3 years ago

Read 2 more answers

Last Saturday jay spent 6 hours playing games,studying, and talking with friends. He spent 2/5 of the time playing and 1/2 of th

LenaWriter [7]

Jay spent 3/7 hours talking with his friends

7 0

3 years ago

What is the number in scientific notation? 409,200,000,000

gavmur [86]

4.092 x 10 to the 11th power

7 0

3 years ago

Read 2 more answers

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 $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .