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2 years ago

HELPPP ILL GIVE 70 POINTS!!!!!

Mathematics

1 answer:

MrRissso [65]2 years ago

6 0

Answer:

It will take 4.71 hours to travel a distance of 221.43 miles

Step-by-step explanation:

Given

Distance covered in time 7.99 hours is 375.32 miles.

Now we know that speed is equal to distance per unit time.

Therefore,

$speed = \frac{distance}{time} \\= \frac{375.32}{7.99}\\=46.97\ mi/h$

Which is the constant speed, hence now time taken for a distance of 221.43 miles will be

$time = \frac{distance}{speed} \\=\frac{221.43}{46.97} \\=4.71\ h$

Therefore, the time taken to cover 221.43 miles is 4.71 hours.

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Complete the equation of the line through (2,1) and (5,-8)

Mila [183]

First we calculate the slope (which is called "m".

Slope = m = (Y2 -Y1) ÷ (X2 -X1)

m = (-8 -1) / (5 -2) = -9 / 3 = -3

We next fill in this equation: b = y -mx

We know m = -3 and we can choose a point (2, 1) to calculate "b"

b = 1 --3 * 2 b = 7

Putting this into y = mx +b

y = -3x +7

7 0

2 years ago

What is this equation<br> 6(14+8)+12

Bess [88]

add 14+8 first, which is 22

then, multiply 6 by 22, which is 132

and lastly, add 12

the answer is 144

5 0

3 years ago

What’s the product of....<br><br> (0.25)•(1.4)=

wolverine [178]

Answer:

0.35

Step-by-step explanation:

0.25 × 1.4 =

0.25×100=25

1.4 × 10 = 14

25×10=250

25×4=100

250+100=350

25×14=350

350÷1000(0.25》25, 1.4》14)

0.35

4 0

3 years ago

The height of a soccer ball that is kicked from the ground can

AveGali [126]

Answer:

87.5

Step-by-step explanation:

By completing the square,

=-14x^2+70x

=-14(x^2-5x)

=-14(x-5/2)^2+175/2

the max. height=175/2

i.e. 87.5

8 0

2 years ago

Plzzz help me with this I’ll give brainliest

scoray [572]

Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-(-1)}{(-4)-(-4)}$

m= $\frac{7}{0}$

The slope is 90 degree

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-6}{(-4)-(-1)}$

m= $\frac{0}{(-3)}$

The slope is zero degree

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-4)-6}{2-2}$

m= $\frac{-10}{0}$

The slope is 90 degree

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-(-4)}{(-4)-(2)}$

m= $\frac{3}{-6}$

m= $\frac{-1}{2}$

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{2-3}{3-(-2)}$

m= $\frac{-1}{5}$

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-2}{2-3}$

m= $\frac{-3}{-1}$

m=3

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{0-(-1)}{(-3)-2}$

m= $\frac{-1}{5}$

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{3-0}{(-2)-(-3)}$

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

3 0

3 years ago