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DedPeter [7]
2 years ago
7

Please help me!!! some easy algebra

Mathematics
2 answers:
Vadim26 [7]2 years ago
6 0

I think this answer is correct We have read this question like this only

kow [346]2 years ago
4 0

Answer:

5xy-x^2t+2x7+3x^2t= 7xy+2x^2t

Step-by-step explanation:

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Solve for x.
konstantin123 [22]

Answer:

The value of x = 9

Step-by-step explanation:

Angle Bisector Theorem

An angle bisector of a triangle would cut the opposite side into two segments that are proportional to the other two sides of the triangle.

Thus,

\frac{x}{15}=\:\frac{12}{20}

Cross Multiply the expression

\:20\times x=12\times 15

Divide both sides by 20

\frac{20x }{20}=\frac{180}{20}

x=9

Therefore, the value of x = 9

5 0
2 years ago
in all he walks a total 7/8 of a mile to and from school each day. How many miles does he walk in one way ?
solmaris [256]

Answer:

The answer is 1.75 miles.

Step-by-step explanation:

5 0
2 years ago
What is the area of a triangle with base 12.5 m and height 2.4 m
BaLLatris [955]
A = 1/2 (b x h)
A = 1/2(12.5 x 2.4)
A = 1/2(30)
A = 15

3 0
3 years ago
Read 2 more answers
The product 2 x 3/6 is shown by the fraction
Neko [114]

Answer:

2*3/6 (=2/6=3)

3/3 =1

Your answer is 1

3 0
2 years ago
The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 25
Assoli18 [71]

Answer:

(a) The velocity after 2 second is 5.9 m/s

The velocity after 4 second is -13.7 m/s.

(b) The projectile reaches its maximum height  after 2.60 s of projection.

(c)The maximum height that is attained by the projectile is 37.18 m.

(d)Therefore the projectile hits the ground after 5.36 seconds of projection.

(e)The velocity of the projectile when it hits the ground is 27.03 m/s

Step-by-step explanation:

Given that, a projectile shot vertically upward from a point 4 m above the ground with a initial velocity of 25.5 m/s.

The height of the projectile after t seconds is

h=4+25.5t-4.9t^2

where h is in meter.

(a)

We use the formula

v=u+at

V= final velocity

u = initial velocity = 25.5 m/s

a = acceleration=   acceleration due to gravity= 9.8 m/s²

Since the object moves upward direction and acceleration due to gravity is downward direction. So here a= -9.8 m/s.

v(2)= 25.5+(-9.8)×2

     =25.5-19.6

     =5.9 m/s

And when t= 4

v(4)= 25.5+(-9.8)×4

    =25.5-39.2

    = -13.7 m/s

The velocity after 2 second is 5.9 m/s

The velocity after 4 second is -13.7 m/s.

(b)

At its maximum height,the velocity of the projectile is zero. i.e v=0

∴0=25.5+(-9.8)t

⇒9.8t=25.5

\Rightarrow t=\frac{25.5}{9.8}

⇒t = 2.60 s

The projectile reaches its maximum height  2.60 s after projection.

(c)

To find the maximum height, we are putting t= 2.60 in this equation h=4+25.5t-4.9t^2.

\therefore h= 4+(25.5\times 2.60)-(4.9\times 2.60^2)

     =37.18 m

The maximum height that is attained by the projectile is 37.18 m.

(d)

When the projectile hits the ground the height will be zero i.e h=0

From the equation of height we get

\therefore h=0=4+25.5t-4.9t^2

\Rightarrow 4+25.5t-4.9t^2=0

\Rightarrow t=\frac{-25.5\pm\sqrt{25.5^2-4(-4.9).4}}{2(-4.9)}

⇒t= -0.15 ,5.36

Therefore it hits the ground after 5.36 seconds of projection.

(e)

To find the velocity we use the formula v=u+at

Here v = final velocity=?

u=25.5 m/s,

t = 5.36 s

a= -9.8m/s²

v=25.5+(-9.8)5.36

 = -27.03 m/s

Negative sign denoted that the motion of the projectile is downward direction.

The velocity of the projectile when it hits the ground is 27.03 m/s.

6 0
2 years ago
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