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3 years ago

Are these answers correct

Mathematics

1 answer:

professor190 [17]3 years ago

7 0

No you have to move the decimal place over 2 times to the left

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Find the measure of the indicated angle to the nearest degree. Thanks.

Sholpan [36]

Answer:

θ ≈ 40°

Step-by-step explanation:

Since, sinθ = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

cosθ = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

tanθ = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

In the picture attached,

Measures of adjacent side and opposite side of the triangle have been given. Therefore, tangent rule will be applied in the given triangle.

tanθ = $\frac{19}{23}$

θ = $\text{tan}^{-1}(\frac{19}{23})$

θ = 39.56

θ ≈ 40°

5 0

2 years ago

Marcel made an error when he used the distributive property to solve the problem shown. Which sentence describes the error?

BigorU [14]

A.In Step 2, 90 should be multiplied by the quantity 10 – 12, not by 10.

correct:
90(10 - 12) = 90 · 10 - 90 · 12 or 90(10 - 12) = 90 · (-2)

7 0

2 years ago

Read 2 more answers

If you were asked to find the total number of tiles covering a circular floor, would you look to find the area or circumference?

vagabundo [1.1K]

Answer:

The area.

Step-by-step explanation:

The circumference would only cover an outline of the floor. Hope this helps!

3 0

2 years ago

PLEASE HURRY!!!!!!!<br> 1) Find the 8th term of the sequence. <br> 6, 12, 24, 48, 96,.......

RideAnS [48]

It would be 768 because you just keep multiplying by 2.

6 0

3 years ago

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Find all real and imaginary solutions to the equation.<br> x^3-2x^2+16x-32=0

777dan777 [17]

Answer:

The solution of $x^3-2x^2+16x-32=0$ is $\mathbf{x=2,x=4i,x=-4i}$

So, the real solutions is: x = 2

Imaginary solutions are : x = 4i, x = -4i

Step-by-step explanation:

We need to find real and imaginary solutions to the equations $x^3-2x^2+16x-32=0$

First of all we will make groups

$(x^3-2x^2)(+16x-32)=0$

Finding common terms from the groups

$x^2(x-2)+16(x-2)=0\\(x-2)(x^2+16)=0$

Now we know that if ab=0 then a=0 , b=0

$x-2=0, x^2+16=0\\Simplifying:\\x=2, x^2=-16\\Taking\: square\: root\\x=2,\sqrt{x^2}=\sqrt{-16}\\x=2, x=\pm\sqrt{-16}\\x=2,x=\m\sqrt{-1}\sqrt{16} \\We\:know\:\sqrt{-1}=i \:and \sqrt{x} \:\sqrt{16}=4 \\x=2,x=\pm4i\\x=2,x=4i,x=-4i$

The solution of $x^3-2x^2+16x-32=0$ is $\mathbf{x=2,x=4i,x=-4i}$

So, the real solutions is: x = 2

Imaginary solutions are : x = 4i, x = -4i

3 0

2 years ago