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3 years ago

PLZ HELP ME!! :D What is the quotient of 2,173 ÷ 53?

Mathematics

2 answers:

Nuetrik [128]3 years ago

6 0

Hello!

I believe the sum of 2,172÷ 53 = 41.

I hope it helps!

exis [7]3 years ago

4 0

its 41. I hope this helps

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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

Ill mark brainlist plss help

mojhsa [17]

Answer:

w=20

Step-by-step explanation:

Solve for w by cross multiplying.

6 0

3 years ago

Read 2 more answers

Which systems of equations intersect at point A in this graph?

abruzzese [7]

Answer:

The point of intersection of the system of equations is:

(x, y) = (-2, 1)

The correct system of equations intersect at point A in this graph will be:

$\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}$

Thus, the second option is correct.

Step-by-step explanation:

Given the point

A (-2, 1)

Let us check the system of equations to determine whether it intersect at point A in this graph.

Given the system of equations

$\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}$

Arrange equation variable for elimination

$\begin{bmatrix}y-4x=9\\ y+3x=-5\end{bmatrix}$

$y+3x=-5$

$-$

$\underline{y-4x=9}$

$7x=-14$

so the system of equations becomes

$\begin{bmatrix}y-4x=9\\ 7x=-14\end{bmatrix}$

Solve 7x = -14 for x

$7x=-14$

Divide both sides by 7

$\frac{7x}{7}=\frac{-14}{7}$

Simplify

$x = -2$

For y - 4x = 9 plug in x = 2

$y-4\left(-2\right)=9$

$y+4\cdot \:2=9$

$y+8=9$

Subtract 8 from both sides

$y+8-8=9-8$

Simplify

y = 1

Thus, the solution to the system of equations is:

(x, y) = (-2, 1)

From the attached graph, it is also clear that the system of equations intersects at point x = -2, and y = 1.

In other words, the point of intersection of the system of equations is:

(x, y) = (-2, 1)

Therefore, the correct system of equations intersect at point A in this graph will be:

$\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}$

Thus, the second option is correct.

3 0

3 years ago

The product is???<br><br> It is multiplication.

loris [4]

Answer:

123.76

Step-by-step explanation:

-3.4(-5.2) = 17.68

17.68(7)

123.76

Not sure if its correct but hope it helps!

7 0

3 years ago

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As part of a new advertising campaign, a beverage company wants to increase the dimensions of their cans by a multiple of 1.10.

Firdavs [7]

Answer:

New can holds $112.25\,\,cm^3$ more than the old can

Step-by-step explanation:

Given: Diameter of the can is 6 cm and height is 12 cm such that volume of can is $339.12\,\,cm^3$

Dimensions of the can are increased by a multiple of 1.10

To find: Difference between the volume of new can and volume of old can

Solution:

Volume of can (v) = $339.12\,\,cm^3$

Let r, h denote radius and height of the can.

Let R, H denotes radius and height of the new can.

r = diameter/2 = $\frac{6}{2}=3\,\,cm$

h = 12 cm

R = $3(1.1)=3.3.\,\,cm$

H = $12(1.1)=13.2\,\,cm$

New volume (V) = $\pi (R)^2H=\pi(3.3)^2(13.2)=451.37\,\,cm^3$

So,

$V-v=451.37-339.12=112.25\,\,cm^3$

7 0

4 years ago