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maw [93]
3 years ago
8

Help me help me!!!!!!!

Mathematics
1 answer:
neonofarm [45]3 years ago
8 0
The answer is B. -x^3+2x^2-3x+10
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Evaluate the expression (x^5)^3(16x^8)^1/2
Contact [7]

Answer:

8x^23 i think

3 0
3 years ago
An equation is shown below:
lbvjy [14]

Step-by-step explanation:

Step 1: 12x-6= 1

step 2:12x=7

5 0
3 years ago
Please help! odfpfpsdfsdlfsdfnlkfn pleaseeeeee
sammy [17]

Answer:

x = 13

Step-by-step explanation:

Since the two given angles are vertical angles, we know (5x) = 65.

Divide both sides by 5 : <u>5x = 65</u>    and we get x = 13

                                             5

7 0
3 years ago
The coordinates of each rectangle ABCD are A (0,2) B (2,4) C (3,3) D (1,1). Find the distance of each side of the rectangle then
EastWind [94]

AB = CD = √8 ≈ 2.8 units

BC = AD = √2 ≈ 1.4 units

Area of the rectangle ABCD = 3.92 units²

Perimeter of the rectangle ABCD = 8.4 units

<h3>How to Find the Area and Perimeter of a Rectangle?</h3>

Given the coordinates of vertices of rectangle ABCD as:

  • A(0,2)
  • B(2,4)
  • C(3,3)
  • D(1,1)

To find the area and perimeter, use the distance formula to find the distance between A and B, and B and C.

Using the distance formula, we have the following:

AB = √[(2−0)² + (4−2)²]

AB = √[(2)² + (2)²]

AB = √8 ≈ 2.8 units

CD = √8 ≈ 2.8 units

BC = √[(2−3)² + (4−3)²]

BC = √[(−1)² + (1)²]

BC = √2 ≈ 1.4 units

AD = √2 ≈ 1.4 units

Area of the rectangle ABCD = (AB)(BC) = (2.8)(1.4) = 3.92 units²

Perimeter of the rectangle ABCD = 2(AB + BC) = 2(2.8 + 1.4) = 8.4 units

Learn more about the area and perimeter of rectangle on:

brainly.com/question/24571594

#SPJ1

5 0
1 year ago
Show that y=cos(t)y=cos(t) is a solution to (dydt)2=1−y2(dydt)2=1−y2. Enter your answers below in terms of the independent varia
GrogVix [38]

Answer:

(\frac{dy}{dt})^2=sin^2t

Step-by-step explanation:

y=cost

DE :(\frac{dy}{dt})^2=1-y^2

If y is a solution of given DE then it satisfied the DE.

Differentiate w.r.t t

\frac{dy}{dt}=-sint

Using the formula

\frac{d(cosx)}{dx}=-sinx

LHS:(\frac{dy}{dt})^2=(-sint)^2=sin^2t

RHS

1-y^2=1-cos^2t=sin^2t

By using the formula

sin^2t=1-cos^2t

LHS=RHs

Hence, y is a solution of given DE

(\frac{dy}{dt})^2=sin^2t

7 0
3 years ago
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