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Svetlanka [38]
3 years ago
13

What is this answer????

Mathematics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

im pretty sure its c

Step-by-step explanation:

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I need help on this thank you!
Digiron [165]
First off both triangles form a 90° angle there both congruent. you can tell they form a 90° angle because of square box
3 0
3 years ago
Jimmy gets a discount of 35% for a pair of sneakers with an original price of $70. He also gets a discount of 25% on a shirt wit
Tanya [424]
First find the total without using the dicounts. then do 70×.35 which is 35% as a decimal. your answer is the dicounts but that is how much you are taking off. .35×70= 24.5 then you would do 70-24.5=45.5 then do the same for the other amount. add these two amounts together. ( 30×.25=7.5 30-7.5= 22.5) 45.5+22.5= 68 then do your first answer - 68 and that is how much is saved
8 0
3 years ago
Read 2 more answers
Lillian is going to invest in an account paying an interest rate of 5.7% compounded continuously. How much would Lillian need to
Tems11 [23]

Answer: $5156

Step-by-step explanation:

Given

The rate of interest is 5.7%

time t=7 years

Amount after 7 years, A=\$7600

for compound interest, the Amount is given by

\Rightarrow A=P(1+\frac{r}{100})^T

Put values

\Rightarrow 7600=P(1+0.057)^7\\\\\Rightarrow P=\dfrac{7600}{(1.057)^7}=5,155.71\\\\\Rightarrow P\approx \$5156

5 0
3 years ago
The Cartesian coordinates of a point are given. (a) (−3, 3) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0
irina [24]

Answer:

a) (-3, 3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (3√2, 0.75π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-3√2, 1.75π)

b) (4, 4√3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (8, 0.13π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-8, 1.13π)

Step-by-step explanation:

We know that polar coordinates are related to (x, y) coordinates through

x = r cos θ

y = r sin θ

And r = √[x² + y²]

a) For (-3, 3)

(i) x = -3, y = 3

r = √[x² + y²] = √[(-3)² + (3)²] = √18 = ±3√2

If r > 0, r = 3√2

x = r cos θ

-3 = 3√2 cos θ

cos θ = -3 ÷ 3√2 = -(1/√2)

y = r sin θ

3 = 3√2 sin θ

sin θ = 3 ÷ 3√2 = (1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 0.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (3√2, 0.75π)

(ii) When r < 0, r = -3√2

x = r cos θ

-3 = -3√2 cos θ

cos θ = -3 ÷ -3√2 = (1/√2)

y = r sin θ

3 = -3√2 sin θ

sin θ = 3 ÷ -3√2 = -(1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 1.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (-3√2, 1.75π)

b) For (4, 4√3)

(i) x = 4, y = 4√3

r = √[x² + y²] = √[(4)² + (4√3)²] = √64 = ±8

If r > 0, r = 8

x = r cos θ

4 = 8 cos θ

cos θ = 4 ÷ 8 = 0.50

y = r sin θ

4√3 = 8 sin θ

sin θ = 4√3 ÷ 8 = (√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (8, 0.13π)

(ii) When r < 0, r = -8

x = r cos θ

4 = -8 cos θ

cos θ = 4 ÷ -8 = -0.50

y = r sin θ

4√3 = -8 sin θ

sin θ = 4√3 ÷ -8 = -(√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 1.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (-8, 1.13π)

Hope this Helps!!!

8 0
3 years ago
2. Cross out the equations that are NOT linear functions.
wel

Welcome to the concept of the maths.

Here, we can see, that Linear equation always have highest power or degree of its variable as '1' ,

thus here, y = 3/x has degree as -1 and y = x^2 +5 has degree 2 ,

hence,

The equations that are NOT linear are

2.) y = 3/x and 5.) y = x^2 + 5

8 0
3 years ago
Read 2 more answers
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