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3 years ago

14

Please help me if you can!!!!!

1 answer:

klio [65]3 years ago

3 0

Filling in the given value, you have

... (sin θ)² + (3/4)² = 1

... (sin θ)² = 1 - 9/16 = 7/16 . . . . . . . subtract the constant on the left

... sin θ = √(7/16) . . . . . . . . . . . . . . take the square root

... sin θ = (√7)/4

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Please help!<br> "Congruence and Constructions".

marin [14]

Answer:

MNK = TRP

MN is congruent to TR.

X = 6.3

Step-by-step explanation:

MNK is congruent to TRP because the two triangles are congruent, which is a given statement. This also means that all side and angle measures will be congruent. Because of this, it can be inferred that MN is congruent to TR and they have equal side length measures. This means that TR measures 20. This means that 3X - 1 = 20. By solving we can get 6.33.

7 0

3 years ago

2x+5y=10<br><br>x intercept=<br><br>y intercept=

Stells [14]

Answer:

x= 0/-8

y= 0/5

Step-by-step explanation:

Brain-List?

5 0

3 years ago

At a sale, books are sold at $90.75 each, which is 75 percent of the original price. What is the original price

lions [1.4K]

$\frac{75}{100} = \frac{90.75}{x} \\ x = \frac{90.75 \times 100}{75} = \frac{9075}{75} = 121$
the answer is 121

6 0

3 years ago

Please help! Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact valu

Nadya [2.5K]

Answer:

$The \ area \ of \ the \ figure = 4\dfrac{1}{2} \cdot \pi + 18 \ cm^2$

Step-by-step explanation:

Given that the figure is made up of portion of a square and a semicircle, we have;

BC ≅ AB = 6 cm

The area of semicircle BC with radius BC/2 = 3 is 1/2×π×r² = 1/2×π×3² = 4.5·π cm²

Triangle ABC = 1/2 × Area of square from which ABC is cut

The area of triangle ABC = 1/2×Base ×Height = 1/2×AB×BC = 1/2×6×6 = 18 cm²

The area of the figure = The area of semicircle BC + The area of triangle ABC

The area of the figure = 4.5·π cm² + 18 cm² = $\dfrac{9 \cdot \pi +36}{2} \ cm^2$ .

5 0

3 years ago

Identify the data set that could be quadratic. HELP ASAP!!

Ne4ueva [31]

Answer:

Option B can be quadratic.

Step-by-step explanation:

Take a look at the data.

A quadratic equation is of the form ,

$y = ax^{2} +bx + c$ ,

where, a,b,c are constants.

<em>To find an unknown equation of 3 variables,</em>

<em>we need 3 points lying on the equation.</em>

here they have given, 5 points, meaning all should lie on the curve.

<em>For first option,</em> inserting first 3 points to find equation, we get equation as,

$y = x^{2} -4x +11$ , but the rest points don't satisfy the curve.

So first option is not quadratic.

Similarly, it can be shown that option C and D are also not quadratic.

While, in option B, it is clear that for every y, x is y squared,

$x = y^{2}$ , thus quadratic in nature.

6 0

3 years ago