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3 years ago

Find arc YWPlease help

Mathematics

2 answers:

Bas_tet [7]3 years ago

8 0

Answer:

Step-by-step explanation:

i swear

nataly862011 [7]3 years ago

7 0

The answer to this is 84

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Let me know the answer plz

Cloud [144]

Answer: Option G

$a_6 = -2048$

Step-by-step explanation:

The geometric series have the following form:

$a_n = a_1 (r) ^ {n-1}$

Where $a_1$ is the first term of the sequence and r is the common radius.

In this case we know that

$a_1 = -2\\\\r = 4$

So:

$a_n = -2(4) ^ {n-1}$

To find the sixth term $a_6$ , substitute $n = 6$ in the equation.

$a_6 = -2 (4) ^ {6-1}\\\\a_6 = -2 (4) ^ 5\\\\a_6 = -2048$

8 0

3 years ago

Express the polynomial 7x + 3x + 5 in standard form

frosja888 [35]

The equation is already in the standard form.

= 10 x + 5

3 0

3 years ago

(40 POINTS) Which linear function represents this table of values?

lakkis [162]

Answer:

B is the answer

Step-by-step explanation:

take two of the points to plug in and test

point one, when x is 3 y is 9

y=2/3 (3) +7

y=6/3 +7

y= 2 + 7

y=9

test the next point

(9,13)

y=2/3 (9) + 7

y=18/3 +7

y= 6 +7

y=13

4 0

3 years ago

X + 4y = -24 5x + 3y = -35

Salsk061 [2.6K]

Answer:

$(-4,-5 )$

Step-by-step explanation:

You are looking for the point of intersection of these two functions.

Substitution:

This are now your formulas

$\left \{ {{x=-4y-24} \atop {5x+3y=-35}} \right.$

You are going to substitute x, so you can solve for y

$5(-4y-24)+3y=-35$
$-20y-120+3y=-35$
$-17y=85$
$y=-5$

Now you are going to substitute y to find x

$x=-4(-5)-24$
$x=20-24$
$x=-4$

Now lets check if it is correct

$-4+4(-5)=-24 :)$
$5(-4)+3(-5)=-35 :)$

So it is true, making (-4 , -5) your point of intersection.

hope it helps

5 0

3 years ago

Calculate f '(0) for the following continuous function. (If an answer does not exist, enter DNE.)

nordsb [41]

$\bf f(x)=|x|+4\implies f(x)=\sqrt{x^2}+4\implies f(x)=(x^2)^{\frac{1}{2}}+4
\\\\\\
\cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{\underline{2}}(x^2)^{-\frac{1}{2}}\cdot \underline{2} x}\implies \cfrac{dy}{dx}=\cfrac{x}{(x^2)^{\frac{1}{2}}}
\\\\\\
\left. \cfrac{dy}{dx}=\cfrac{x}{\sqrt{x^2}} \right|_{x=0}\implies \stackrel{und efined}{\cfrac{0}{0}}$

3 0

3 years ago

Find arc YWPlease help​

Find arc YWPlease help