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3 years ago

12

Helpppppppp

1 answer:

wel3 years ago

6 0

The hypotenuse can be solved by this formula. X^2 = 6^2 +6^2

X^2=64
Usually you cut it down to 8 but it wants the squared form so 64.

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Find the first four terms for the arithmetic sequence given a1 = 6 and d = 5

DENIUS [597]

Answer:

6, 11, 16, 21

Step-by-step explanation:

To obtain the first 4 terms add the common difference 5 to the previous term, that is

a₁ = 6

a₂ = a₁ + 5 = 6 + 5 = 11

a₃ = a₂ + 5 = 11 + 5 = 16

a₄ = a₃ + 5 = 16 + 5 = 21

7 0

3 years ago

Y= -1/2(x+10)^2+14 Find the range

Andre45 [30]

Answer:

Step-by-step explanation:

y=-1/2(x+10)^2+14

2y=-(x+10)^2+28

(x+10)^2=28-2y

L.H.S. is positive.

so 28-2y≥0

28≥2y

or 2y≤28

or y≤ 14

Range is (-∞,14]

8 0

3 years ago

I Need Help On Multiplying Fractions It’s 5th Grade

nikdorinn [45]

Answer:

Ok ill help

Step-by-step explanation:

7 0

3 years ago

Triangle ABC and DEF are similar what is the length of segment DF?

Ipatiy [6.2K]

Answer:

Proportionate to whatever length the corresponding segment for the similar triangle.

5 0

3 years ago

Pls, can you help me? thx.<br><img src="https://tex.z-dn.net/?f=%20%5Ccos%28x%20%2B%20%5Cfrac%7B%5Cpi%7D%7B3%7D%29%20%5Cgeqslant

Veseljchak [2.6K]

For $x$ between $-\pi$ and $\pi$ , we have

$\cos x=\frac{\sqrt2}2=\frac1{\sqrt2}$ when $x=\pm\frac\pi4$ ;
$\cos x=1$ for $x=0$ ; and
$\cos x=0$ for $x=\pm\frac\pi2$

$\cos x$ is continuous over its domain, so the intermediate value theorem tells us that

$\cos x\ge\frac{\sqrt2}2$

is true for $-\frac\pi4\le x\le\frac\pi4$ .

For all $x$ , we take into account that $\cos x$ is $2\pi$ -periodic, so the above inequality can be expanded to

$-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4$

where $n$ is any integer. Equivalently,

$-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi$

To get the corresponding solution set for

$\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2$

simply replace $x$ with $x+\frac\pi3$ :

$-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi$

$\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}$

7 0

3 years ago