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3 years ago

PLS HELP! I WILL GIVE SOOOOO MANY POINTTSSSS and a brainly!!!

Mathematics

2 answers:

Gnom [1K]3 years ago

5 0

Answer:

the answer to the question would be 7

Allushta [10]3 years ago

5 0

Answer:

Step-by-step explanation:

You have to remember :

125

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I’ll mark Brainly! Andy deposits $25 in his checking account.

Dmitry_Shevchenko [17]

Answer:

1. $25

2. 12

3. -12

4. -12

5. -25

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Part B if the height of each prism is 10 Cm what is the surface area prism

Evgen [1.6K]

Answer:

Prism A:

$Area = 288cm^2$

Prism B:

$Area =250cm^2$

Step-by-step explanation:

Given

See attachment for prisms

$Height(h) = 10cm$

Required

Determine the surface area of both prisms

Prism A is triangular and as such, the surface area is:

$Area = 2 * A_b + (a + b + c) * h$

Where

$A_b = \sqrt{s * (s - a) * (s -b) * (s - c)}$

and

$s = \frac{a + b + c}{2}$

Such that a, b and c are the lengths of the triangular sides of the prism.

From the attachment;

$a = 8; b =6; c =10$

So, we have:

$s = \frac{a + b + c}{2}$

$s = \frac{8 + 6 + 10}{2}$

$s = \frac{24}{2}$

$s = 12$

Also:

$A_b = \sqrt{s * (s - a) * (s -b) * (s - c)}$

$A_b = \sqrt{12 * (12 - 8) * (12 - 6) * (12 - 10)}$

$A_b = \sqrt{576}$

$A_b = 24$

So:

$Area = 2 * A_b + (a + b + c) * h$

$Area = 2 * 24 + (8 + 6 + 10) * 10$

$Area = 288cm^2$

Prism B is a rectangular prism. So, the area is calculated as:

$Area = 2 * (ab + bh + ah)$

From the attachment

$a = b = 5$

$h =10$

So:

$Area =2 * (5 * 5 + 5 * 10 + 5 * 10)$

$Area =250cm^2$

7 0

2 years ago

WILL GIVE BRAINLEST ANSWER IF ANSWERED IN THE NEXT 24 HRS Express the complex number in trigonometric form. -5i

trasher [3.6K]

Answer:

$z=5\left(\cos \left(\dfrac{3\pi}{2}\right)+i\sin \left(\dfrac{3\pi}{2}\right)\right)$

Step-by-step explanation:

If a complex number is z=a+ib, then the trigonometric form of complex number is

$z=r(\cos \theta +i\sin \theta)$

where, $r=\sqrt{a^2+b^2}$ and $\tan \theta=\dfrac{b}{a}$ , $\theta$ is called the argument of z, $0\leq \theta\leq 2\pi$ .

The given complex number is -5i.

It can be rewritten as

$z=0-5i$

Here, a=0 and b=-5. $\theta$ lies in 4th quadrant.

$r=\sqrt{0^2+(-5)^2}=5$

$\tan \theta=\dfrac{-5}{0}$

$\tan \theta=\infty$

$\theta=2\pi -\dfrac{\pi}{2}$ $[\because \text{In 4th quadrant }\theta=2\pi-\theta]$

$\theta=\dfrac{3\pi}{2}$

So, the trigonometric form is

$z=5\left(\cos \left(\dfrac{3\pi}{2}\right)+i\sin \left(\dfrac{3\pi}{2}\right)\right)$

4 0

3 years ago

Read 2 more answers

Which statements are true about the place values in the number 99.999?

allsm [11]

Answers A. and B. Are correct

5 0

3 years ago

PLEASE SOLVE THIS PROBLEM

schepotkina [342]

Answer:

18) Area= 5*5/2=25/2=12.5 unit ^2

19) Area=AB^2V3/4=8a^2*V3/4=2V3a^2

Step-by-step explanation:

18. A(-3,0)

B(1,-3)

C(4,1)

AB=V(-3-1)^2+(0+3)^2=V16+9=V25=5

AC=V(-3-4)^2+(0-1)^2=V49+1=V50=5V2

BC=V(1-4)^2+(-3-1)^2=V9+16=V25=5

so AB=BC=5

and AC^2=AB^2+BC^2

so trg ABC is an isosceles right angle triangle (<B=90)

Area= 5*5/2=25/2=12.5 unit ^2

19. A(a,a)

B(-a,-a)

C(-V3a, V3a)

AB=V(a+a)^2+(a+a)^2=V4a^2+4a^2=V8a^2

AC=V(a+V3a)^2+(a-V3)^2=Va^2+2a^2V3+3a^2+a^2-2a^2V3+3a^2=V8a^2

BC=V(-a+V3a)^2+(-a-V3a)^2=V8a^2

so AB=AC=BC

Area=AB^2V3/4=8a^2*V3/4=2V3a^2

7 0

2 years ago