Please answer this multiple choice question only if you know the answer!! 25 points and brainliest!

Mathematics

1 answer:

solniwko [45]3 years ago

5 0

Answer:

i cant see it

Step-by-step explanation:

You might be interested in

A skate ramp is 4.2 meters long. It rises up 1.7 meters. What’s is the angle of elevation to the nearest tenth of a degree?

Kipish [7]

Θ
=
arcsin
(
.7
4.2
)
≈
10
∘
Explanation:
We view the ramp as a right triangle. The hypotenuse is 4.2 and the vertical side .7, which is opposite the angle
θ
we seek.
sin
θ
=
.7
4.2
=
1
6
I'm going to finish the problem but I'll note if we were actually building the ramp we don't need to know the angle; this sine is sufficient.
θ
=
arcsin
(
1
6
)
θ
≈
10
∘
which I think is a pretty steep ramp for a wheelchair.
There will be another inverse sine that is the supplementary angle, around
170
∘
, but we can rule that out as a value for a ramp wedge angle.

4 0

3 years ago

Solve for x, rounding to the nearest hundredth. 10^3x = 98

Anna71 [15]

Answer:

see explanation

Step-by-step explanation:

10^3x=98

1000x=98

x=98/1000

x=49/500

x=0.98

if it is rounded to the nearest hundredth then x=0

8 0

3 years ago

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$