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yuradex [85]
4 years ago
13

Help my friend out she dumb

Mathematics
1 answer:
garik1379 [7]4 years ago
5 0

Answer:

I'm dumb too, but i think its 12....

Step-by-step explanation:

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UJUU J
Stels [109]

Answer:

Plese read the complete procedure below:

Step-by-step explanation:

The polynomial is p(a) = (a^4 - 6a^3 + 3a^2 + 26a – 24)

a)

1   -6   3   26   -24    |<u>  1     </u>

<u>      1   -5   -2    24</u>

1   -5   -2   24    0

The remainder is zero, then (a-1) is a factor of the polynomial

b)

1   -6   3   26   -24    |<u>  2     </u>

<u>      2  -8   10    72</u>

1   -4   5   36    48

When p(a) is divided by (a-2) the remainder 28/p(a)

1   -6   3   26   -24    |<u> - 4     </u>

<u>    -4  40  172  -792</u>

1   -10  43  198  -816

When p(a) is divided by (a-2) the remainder -816/p(a)

c) I attached an image of the long division below:

3 0
3 years ago
Use the net to find the lateral area of the cylinder. Give your answer in terms of pi.
MatroZZZ [7]

Answer:

90\pi in^2

Step-by-step explanation:

The lateral  area of the cylinder is given by;

A_L=2\pi rh

The radius of the cylinder is r=5in. and the height is r=9in.

Substitute the values to get;

A_L=2\pi \times 5\times 9

A_L=90\pi in^2

5 0
4 years ago
Read 2 more answers
A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, i
Cerrena [4.2K]

Answer:

a.2s b.43.6m

Step-by-step explanation:

h(t)=−4.9t2+19.6t+24

24m = inittial possition

-4.9 t2= -1/2 g t2

19.6t=  Vo t

as the initial velocity is positive, the ball is thrown up.

the maximum height of the ball is reached when the velocity is 0

V= Vo-gt=19.6m/s-9.8m/s2 t=0

t=19.6/9.8=2s

<u>it takes 2seconds for the ball to reach its maximum height</u>

h(t)=−4.9t2+19.6t+24

h(2)=-19.6m+39.2m+24=43.6m

<u>the maximum height of the ball is 43.6m</u>

5 0
3 years ago
Helppp!!!!
-Dominant- [34]

Answer:

30

Step-by-step explanation:

Let's say x is the number of friends he has

4 · x = 121 - 1

4x = 120

Divide by 4 on each side

x = 30

7 0
3 years ago
2)If triangle RPQ is a right angled triangle at Q. If PQ = 5cm and RQ = 10 cm,find (i) sin ^2 P (ii) cos ^2 R and tan R (iii) si
Scorpion4ik [409]

Answer:  (i) 1       (ii) 3/4 , √3/3       (iii) 0       (iv) 1

<u>Step-by-step explanation:</u>

ΔRPQ is 1 30°-60°-90° triangle <em>because RQ = 2PQ</em> , where

  • ∠R = 30°
  • ∠P = 90°
  • ∠Q = 60°

(i) sin² (90°) = 1² = 1

\text{(ii)}\ \cos ^2(30^o)= \bigg(\dfrac{\sqrt3}{2}\bigg)^2=\dfrac{3}{4}

.\quad \tan (30^o)=\dfrac{1}{\sqrt3}\bigg=\dfrac{\sqrt3}{3}

(iii) sin (90°) × cos (90°) = 1 × 0 = 0

(iv) sin² (90°) - cos² (90°) = 1² - 0² = 1

5 0
3 years ago
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