(sin30°+cos30°)-(sin60°cos60°) find the value

Quiz! <br>2x + 2 = 4 <br>find x! <br>

lilavasa [31]

<h3>Solution</h3>

2x + 2 = 4
2x = 4 - 2
2x = 2
x = 2 ÷ 2
x = 1

<h3>Proof</h3>

2(1) + 2
= 2 + 2
= 4

7 0

3 years ago

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What is the range of the relation below?

Greeley [361]

[6, -infinity)

have a good night

5 0

3 years ago

Theorem: The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. A two-c

natali 33 [55]

Answer: By the slope formula.

Step-by-step explanation:

Given: ABC is a triangle (shown below),

In which A≡(6,8), B≡(2,2) and C≡(8,4)

And, D and E are the mid points of the line segments AB and BC respectively.

Prove: DE║AC and DE = AC/2

Proof:

Since, And, D and E are the mid points of the line segments AB and BC respectively.

Therefore, By mid point theorem,

coordinate of D are $(\frac{2+6}{2} , \frac{2+8}{2} ) = (\frac{8}{2} , \frac{10}{2} )= (4,5)$

Coordinate of E are $(\frac{2+8}{2} , \frac{2+4}{2} ) = (\frac{10}{2} , \frac{6}{2} )= (5,3)$

By the distance formula,

$DE=\sqrt{(5-4)^2+(3-5)^2}=\sqrt{5}$

$AC=\sqrt{(8-6)^2+(4-8)^2}=2\sqrt{5}$

By the slope formula,

Slope of AC = $\frac{4-8}{8-6} = \frac{-4}{2} = -2$

Slope of DE = $\frac{3-5}{5-4} = \frac{-2}{1} = -2$

Statement Reason

1. The coordinate of D are (4,5) and 1. By the midpoint formula

the coordinate of E are (5,3)

2. The length of DE = √5 2. By the Distance formula

The length AC = 2√5 ⇒ Segment DE

is half the length of segment AC

3. The slope of DE = -2 and the 3. By the slope formula

slope of AC = -2

4. DE║AC 4. Slopes of parallel lines are equal

7 0

3 years ago

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F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

4 0

3 years ago

The triangle STU shown on the coordinate grid below is reflected once to map onto triangle S'T'U': Triangle STU drawn on a 4 qua

Andreyy89

Given that triangle <span>STU is reflected once to map onto triangle S'T'U'.
Given that triangle STU has vertices S(8, 6), T(2, 2), U(5, 1).

If vertex T' is at (2, −2), this means that triangle STU is refrected across the x-axis.

A refrection across the x-axis results in an image that has the same x-value as the pre-image but a y-value that has the opposite sign of the y-value of the pre image.

Thus, a point, say (x, y), refrected over the x-axis will result in an image with coordinate (x, -y)

Therefore, given that the coordinate of S is (8, 6), then the coordinates of vertex S'</span> is (8, -6).

4 0

3 years ago

(sin30°+cos30°)-(sin60°cos60°) find the value​

(sin30°+cos30°)-(sin60°cos60°) find the value