All categories

3 years ago

Can pls someone help and show me how to solve i’m really struggling

Mathematics

2 answers:

a_sh-v [17]3 years ago

7 0

Answer:

156

Step-by-step explanation:

The angles are a linear pair so that means the sum should add up to 180 degrees.

RQS + YQS = 180

We already know that TQS = 24

RQS + 24 = 180

x + 24 = 180

x = 156

Hope this helps :)

Nadya [2.5K]3 years ago

5 0

Answer:

∠RQS = 156

Step-by-step explanation:

m∠TQS = 24

∠TQS and ∠RQS are linear pair. Linear pair are two adjacent angles and their sum is 180

∠TQS + ∠RQS = 180

24 + ∠RQS = 180

∠RQS = 180 - 24

∠RQS = 156

You might be interested in

Help me work this out please

creativ13 [48]

Answer:

∠DEF = 250°

Step-by-step explanation:

1. This shape is a hexagon (6 sides), so the interior angles should all add to 720°. Also worth noting: the problem says the <em>obtuse</em> angle DEF; this means it's the angle INSIDE the shape, not outside.

2. Add all the known angles, then subtract from the total degrees:

50 + 96 + 144 + 42 = 332

720 - 332 = 338

3. Because BC is parallel to ED, we can subtract 180 - 42 for the value of ∠EDC, which is 138°.

4. Add all the known values and subtract from 720 for the value of ∠DEF:

50 + 96 + 144 + 42 + 138 = 470

720 - 470 = 250°

6 0

3 years ago

GRAVITY The height above the ground of a ball thrown up with a velocity of 96 feet per second from a height of 6 feet is 6+96t-1

Nata [24]

Here the time function is h(t) = [6 + 96t - 16t^2] feet.

The initial height of the ball is 6 feet. That's when t=0. h(0)=[6+0-0] ft = 6 ft.

At t=7 sec, h(t) = [6 + 96t - 16t^2] feet becomes
h(7 sec) = h(t) = [6 + 96(7) - 16(7)^2] feet This produces a large negative number (-106 ft), which in theory indicates that the ball has fallen to earth and burrowed 106 feet into the soil. Doesn't make sense.

Instead, let t=1 sec. Then h(1 sec) = h(t) = [6 + 96(1) - 16(1)^2] feet
=[6 + 96 -16] ft, or 86 ft.

One sec after the ball is thrown upward, it reaches a height of 86 feet. It continues to rise, slowing down, until it finally stops for an instant and then begins to fall towards earth.

6 0

4 years ago

What is the Area of this Parallelogram?

NeX [460]

Answer:

C. 134.54

Step-by-step explanation:

b*h = 19.22*7

4 0

3 years ago

How do i fid all missing sides and angles - Find ALL missing sides and angles in the triangle below.

Anarel [89]

Answer:

All three pairs of corresponding sides are equal. Two triangles. Each side of the first triangle is congruent to one side of the ... Two pairs of corresponding sides and the corresponding angles between them ... Two triangles have one congruent angle. ... The SSS similarity criterion allows us to calculate missing side lengths

Step-by-step explanation:

4 0

4 years ago

I honestly need help with these

Brilliant_brown [7]

9. The curve passes through the point (-1, -3), which means

$-3 = a(-1) + \dfrac b{-1} \implies a + b = 3$

Compute the derivative.

$y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}$

At the given point, the gradient is -7 so that

$-7 = a - \dfrac b{(-1)^2} \implies a-b = -7$

Eliminating $b$ , we find

$(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}$

Solve for $b$ .

$a+b=3 \implies b=3-a \implies \boxed{b = 5}$

10. Compute the derivative.

$y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6$

Solve for $x$ when the gradient is 2.

$x^2 - 5x + 6 = 2$

$x^2 - 5x + 4 = 0$

$(x - 1) (x - 4) = 0$

$\implies x=1 \text{ or } x=4$

Evaluate $y$ at each of these.

$\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}$

$\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}$

11. a. Solve for $x$ where both curves meet.

$\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5$

$\dfrac{x^3}3 - 2x^2 - 9x = 0$

$\dfrac x3 (x^2 - 6x - 27) = 0$

$\dfrac x3 (x - 9) (x + 3) = 0$

$\implies x = 0 \text{ or }x = 9 \text{ or } x = -3$

Evaluate $y$ at each of these.

$A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}$

$B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}$

$C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}$

11. b. Compute the derivative for the curve.

$y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8$

Evaluate the derivative at the $x$ -coordinates of A, B, and C.

$A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}$

$B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}$

$C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}$

12. a. Compute the derivative.

$y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}$

12. b. By completing the square, we have

$12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4$

so that

$\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}$

13. a. Compute the derivative.

$y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}$

13. b. Complete the square.

$3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3$

Then

$\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}$

5 0

2 years ago