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3 years ago

14

If SLDG is an isosceles trapezoid, what is the value of x?

1 answer:

MaRussiya [10]3 years ago

8 0

Answer:

13

Step-by-step explanation:

You might be interested in

Hich correctly describes how the graph of the inequality 4x − 2y > 20 is shaded?

joja [24]

Answer:

First we need to express y in this inequality:

-2y > -4x + 20

-y > -2x + 10

y < 2x - 10

The line isn't solid because the "less" sign doesn't include equal in it.

Because the sign is "less than" the shaded part is Below the line.

Merging these 2 conclusion we get the answer:

below the dashed line.

Step-by-step explanation:

6 0

3 years ago

Given the function, f(x)= 1/2x − 4, evaluate f(2).

Norma-Jean [14]

Answer:

-3

Step-by-step explanation:

1/2x2=1 1-4=-3

4 0

3 years ago

Alaina has $28 in her account.She wants to purchase a pair of shoes that costs $45.If Alaina makes the purchase,which integer wi

ki77a [65]

Answer:

-17

Step-by-step explanation:

28 - 45 = -17$ in her account

5 0

3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

7 0

3 years ago

Emma tried to write and equation of the linear table but made an error

Lady bird [3.3K]

Answer:

The correct answer is (-3,-1) (0,5) (3,11)

Step-by-step explanation:

Emma's mistake was flipping her y-values, 11 matches with 3, not -3. The most likely reason she did this was accidentally making the slope negative. The table she wrote fits the line y=-2x+5. If you plug in -3 into the correct equation you will get -1 and 3 will get you 11.

4 0

3 years ago

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