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Anastaziya [24]
3 years ago
8

PLs answer correctly will mark brainliest, 40 points

Mathematics
1 answer:
irakobra [83]3 years ago
6 0

Answer:

  • A, C, A

Step-by-step explanation:

<u>Use properties</u>

  • (mᵃ)ᵇ = mᵃᵇ
  • mᵃ*mᵇ = mᵃ⁺ᵇ

10.

  • (-t⁴)³ = -t⁴ˣ³ = -t¹²
  • Option A

11.

  • (t⁸)² = t⁸ˣ² = t¹⁶
  • Option C

12.

  • (x⁻³)⁻⁵x⁶ = x⁽⁻³⁾ˣ⁽⁻⁵⁾x⁶ = x¹⁵x⁶ = x¹⁵⁺⁶ = x²¹
  • Option A
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Sliva [168]

Answer:

D)

Step-by-step explanation:

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3 years ago
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Joseph's age is three times that of his sister, Isabella. When you add Joseph's age to Isabella's age, you get 36. How old is Jo
kogti [31]
A: x (isabella)
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3 0
3 years ago
I really need help rn T_T plz help me out!
pshichka [43]

Answers:

  • The lengths of sides PQ and RS are <u>   13   </u>
  • The lengths of sides QR and SP are <u>   </u><u>20  </u>

This is a 13 by 20 rectangle.

============================================================

Explanation:

Refer to the drawing below.

Let x be the length of side SP. Since we're dealing with a rectangle, the opposite side is the same length. Side QR is also x units long.

We're told that RS = SP - 7 which is the same as saying RS = x-7

We also know that PQ = x-7 as well because PQ is opposite side RS.

In short, we have these four sides in terms of x

  • PQ = x-7
  • QR = x
  • RS = x-7
  • SP = x

as shown in the drawing. The four sides add up to the perimeter of 66.

PQ+QR+RS+SP = perimeter

PQ+QR+RS+SP = 66

(x-7)+x+(x-7)+x = 66

4x-14 = 66

4x = 66+14

4x = 80

x = 80/4

x = 20

Use this x value to find the unknown side lengths.

  • PQ = x-7 = 20-7 = 13
  • QR = x = 20
  • RS = x-7 = 20-7 = 13
  • SP = x = 20

In short, this is a 13 by 20 rectangle.

-----------------

Check:

perimeter = side1+side2+side3+side4

perimeter = PQ+QR+RS+SP

perimeter = 13+20+13+20

perimeter = 33+33

perimeter = 66

The answer is confirmed.

4 0
2 years ago
Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
Please help. It’s due today
vladimir1956 [14]

Answer:

There's nothing there I believe you forgot to add a link just add or create another question and i'll see what I can do :)

Step-by-step explanation:

5 0
2 years ago
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