Plzz help

Mathematics

1 answer:

Lina20 [59]2 years ago

5 0

Answer:

$14

Step-by-step explanation:

40% of 10 is 4

4+10=14

hope this helps

You might be interested in

Model subtraction of fraction:

cricket20 [7]

The entire race is 1 full race.
One full race is 1.
He ran 3/8 of 1, so he ran 3/8.
He needs to run the rest of the race.
The rest is unknown, so we call it x.
When you add 3/8 to the unknown, you get the full race.
The equation is

x + 3/8 = 1

Change 1 to a denominator of 8.

x + 3/8 = 8/8

Subtract 3/8 from both sides.

x = 5/8

Answer: He still needs to run 5/8 of the race.

5 0

3 years ago

Write a problem that involves finding the unknown measure of a side of a rectangle include the solution?

tankabanditka [31]

Problem:
The area of the patio of a relative's house is 20 square meters.
The width of the patio is 2 meters.
What is the length of the patio?
Solution:
The area is:
A = (w) * (l)
Where,
w: width
l: long
Clearing l we have:
l = A / w
Substituting values:
l = 20/2
l = 10 meters

7 0

2 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

Slope:

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

Slope:

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$