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2 years ago

HELP PLEASE!! I WILL GIVE BRAINLIEST

Mathematics

1 answer:

NISA [10]2 years ago

6 0

77 feet I according to my calculations :)

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Use the order of operations to evaluate this expression: 7 + (5 – 9)2 + 3(16 ÷ 8).

Jet001 [13]

Remember to use PEMDAS when solving.

= 7 + (5 – 9)2 + 3(16 ÷ 8)

= 7 + (-4)2 + 3(2)

= 7 + (-8) + 6

= -1 + 6

= 5

Best of Luck!

3 0

3 years ago

Help please!

Lemur [1.5K]

Answer:

Step-by-step explanation:

a I’m pretty sure

5 0

2 years ago

Read 2 more answers

If QRST is a kite, find mZQRS. R (11x - 6) 79° S (4x + 13) T mZQRS<br>I really need help!

maxonik [38]

Step-by-step explanation:

4x +13 +79+79+ 11x-6 = 360

15x +165 = 360

15x = 195

x =13

QRS = 11(13)-6 = 137

4 0

2 years ago

P and q are two arithmetic sequences. The first four terms of sequence P are 2,6,10,14. The nth term of sequence Q is 145-3n. Th

Tomtit [17]

Answer:

r = 21

Step-by-step explanation:

nth of sequence P :

$T_{n} = 4n -2$

$r^{th} term$ : $4r-2=145-3r\\4r+3r=145+2\\7r=147\\r=21$

5 0

1 year ago

Need help with Calculus 1 inverse trig functions

lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

$\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}$

Where u is a function of x as provided:

$y=3tan^{-1}(x+\sqrt{1+x^2})$

If we set

$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

$\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}$

Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

8 0

2 years ago