All categories

2 years ago

Plz help ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Mathematics

2 answers:

Fantom [35]2 years ago

5 0

Answer:

Step-by-step explanation:

bonufazy [111]2 years ago

4 0

Answer:

Step-by-step explanation:

Lateral Area formula: ph (perimeter x height)

You might be interested in

Which statement best illustrates using the vertical line test to determine if the graph below is a function of x? On a coordinat

Anna35 [415]

Answer:

ans is part a

Step-by-step explanation:

means 1 part

4 0

3 years ago

Read 2 more answers

Verify the identity algebraically:<br> Csc(-x)tanx =-secx

weqwewe [10]

Step-by-step explanation:

Recall that

$\sin(-x) = -\sin x$

Therefore,

$\csc(-x) = \dfrac{1}{\sin(-x)} = -\dfrac{1}{\sin x}$

$\csc(-x)\tan x = \left(-\dfrac{1}{\sin x}\right)\left(\dfrac{\sin x}{\cos x}\right)$

$\:\:\:\:\:\:\:\:\:= -\dfrac{1}{\cos x} = -\sec x$

5 0

2 years ago

Please help me find the solution

tangare [24]

Answer:

Step-by-step explanation:

BD bisects <ABC, this means the half-line divided the angle into two equal parts.

If m<ABC is equal to 44 then m<ABD is

44/2 = 22

4 0

2 years ago

If jerry runs 1\10 mile each day how many days will it take him to run 4\5 mile plz send walkthrough

guajiro [1.7K]

8 days because you multiply 1/10 x 8 = 8/10 after you simplify it you get 4/5 Hope it helped

6 0

2 years ago

Read 2 more answers

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