All categories

2 years ago

Jordan earns 5 dollars

Mathematics

1 answer:

andrew11 [14]2 years ago

4 0

Ok? Is there like an equation or something that I can see so then I can help you?

You might be interested in

Please help me now plz <br> I will mark you brainliest

jek_recluse [69]

Answer:

The vertex angle is 180-(60+75)=45.

k/sin75=900/sin45 so k=900sin75/sin45=1229km approx.

Step-by-step explanation:

6 0

3 years ago

HELP any1 good at math?

valentina_108 [34]

Step-by-step explanation:

the equations are :-

=》d = 40m + 200

=》d = 50m

here, by equalising both equations.

( since, d = d )

=》40m + 200 = 50m

=》50m - 40m = 200

=》10m = 200

=》m = 200 ÷ 10

=》m = 20

now,

=》d = 50m

=》d = 50 × 20

=》d = 1000

6 0

3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

5 0

3 years ago

10,13,14,17,22,25 find the mean

grin007 [14]

Median = 15.5

Please take a look at the image attached for an explanation.

3 0

3 years ago

Read 2 more answers

Last Math problem, I need super extra help solving this math problem. 3y-4-6-3x-7y+6x+1

pantera1 [17]

Answer:

3x-4y-9

Step-by-step explanation:

Combine like terms

5 0

2 years ago