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2 years ago

Can someone check my answers an see if correct. TYSM!! ΔEFG ≅ ΔHJK

Mathematics

1 answer:

trapecia [35]2 years ago

5 0

Answer:

Is here a pic that goes with it?

Step-by-step explanation:

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The intersection of lines y=-14x+3 and y=-32x+3 is .

Kamila [148]

The intersection of lines y=-14x+3 and y=-32x+3 is (0, 3)

<h3>How to determine the intersection of the lines?</h3>

The lines are given as:

y = -14x + 3 and y = -32x+3

Substitute y = -32x+3 in

-32x+3 = -14x + 3

Evaluate the like terms

-18x = 0

Divide by -18

x = 0

Substitute x = 0 in y = -14x + 3

y = -14(0) + 3

Evaluate

y = 3

Hence, the intersection of lines y=-14x+3 and y=-32x+3 is (0, 3)

Read more about linear equations at:

brainly.com/question/14323743

#SPJ1

6 0

2 years ago

A triangle has an area of 56 square units it's height is 14 units

kobusy [5.1K]

The formula for the area of a triangle is 1/2*b*h=a.
1/2*b*14=56
b*14=112
b=8

7 0

3 years ago

HELP ASAP!!!! Does someone mind walking me through this. I really need help.

Tamiku [17]

The minimum of this graph is the focus of the parabola. I'm not sure with the maximum though but I think it doesn't have a maximum because the y value of the parabola will extend infinitely upward.

7 0

3 years ago

11.) Mrs. Collins is buying supplies for her classroom. Pencils cost

ki77a [65]

Answer:

the total cost will be $14

Step-by-step explanation:

a) 0.05p + 0.45f

b) 0.05(100) + 0.45(20)=

5+9

4 0

2 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

5 0

3 years ago