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Rasek [7]
3 years ago
5

Which of the following is the equation of a circle with center (5, -2) and a radius of 3?

Mathematics
1 answer:
ryzh [129]3 years ago
8 0
EQUATION OF A CIRCLE = (x-h)² + (y-k)² =r²

Since the center of the cercle is in O, then h=5 & k=-2

(x-5)² + (y+2² =9
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Two base angles of an isosceles triangle measure (3x + 2) and (5x – 30) degrees. How many degrees are in the third angle of the
joja [24]

Answer:

Step-by-step explanation:

Note that the base angle of an isosceles triangle is equal hence;

3x+2 = 5x-30

3x - 5x = -30-2

-2x = -32

x = -32/-2

x = 16

Hence one of the base angle = 3(16)+2 = 50degrees

The sum of angle in the triangle is 180, hence

50+50+y = 180

100 + y = 180

y = 80 degrees

Hence the third angle is 80 degrees

6 0
3 years ago
First Answer is the brainlest 15 points
KIM [24]

Answer:

−7⋅(−3)⋅(−5)⋅(−8)

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Help me with this question <br> Please I need help
34kurt

Answer:

384 cm²

Step-by-step explanation:

The shape of the figure given in the question above is simply a combined shape of parallelogram and rectangle.

To obtain the area of the figure, we shall determine the area of the parallelogram and rectangle. This can be obtained as follow:

For parallelogram:

Height (H) = 7.5 cm

Base (B) = 24 cm

Area of parallelogram (A₁) =?

A₁ = B × H

A₁ = 24 × 7.5

A₁ = 180 cm²

For rectangle:

Length (L) = 24 cm

Width (W) = 8.5 cm

Area of rectangle (A₂) =?

A₂ = L × W

A₂ = 24 × 8.5

A₂ = 204 cm²

Finally, we shall determine the area of the shape.

Area of parallelogram (A₁) = 180 cm²

Area of rectangle (A₂) = 204 cm²

Area of figure (A)

A = A₁ + A₂

A = 180 + 204

A = 384 cm²

Therefore, the area of the figure is 384 cm²

3 0
3 years ago
Find a linear approximation of the function f(x)=\sqrt[3]{1+x} at a=0, and use it to approximate the numbers \sqrt[3]{.96} and \
34kurt
f(x)=\sqrt[3]{1+x}=(1+x)^{1/3}\implies f'(x)=\dfrac1{3(1+x)^{2/3}}

The linear approximation to f(x) around x=a is then

L(x)=f(a)+f'(a)(x-a)\approx f(x)

So the approximation centered at a=0 will be

L(x)=f(0)+f'(0)x=(1+0)^{1/3}+\dfrac x{3(1+0)^{2/3}}
L(x)=1+\dfrac x3

which means we have

\sqrt[3]{0.96}\approx L(0.96)=1+\dfrac{0.96}3=1.32

\sqrt[3]{1.02}\approx L(1.02)=1+\dfrac{1.02}3=1.34

Compare to the actual values of

\sqrt[3]{0.96}\approx0.9864

\sqrt[3]{1.02}\approx1.0066
4 0
3 years ago
2.184÷0.7 pls help STAT
Tju [1.3M]

Answer:

3.12

Step-by-step explanation:

8 0
3 years ago
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