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astraxan [27]
2 years ago
15

What is the product?

Mathematics
1 answer:
jekas [21]2 years ago
3 0

Answer:

C

Step-by-step explanation:

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I neeeddd help pleaase
Norma-Jean [14]
I got 65 for my answer. Hope that helps!
7 0
3 years ago
4x^2-7x=15<br> Solve by factoring and please please please explain or show the work
vodomira [7]

Answer:

x=3 and x=-1.25

Step-by-step explanation:

4x^2-7x=15

subtract 15 from each side:

4x^2-7x-15=0

factor  (trying different numbers that the sum is 7 and the quotient is 15)

(2x-6 )(2x+2.5 )=0

2x-6=0 add 6 to each side

2x=6

x=3

2x+2.5=0   subtract 2.5 from each side

2x=-2.5

x=-1.25

8 0
3 years ago
Find the​ p-value for the indicated hypothesis test with the given standardized test​ statistic, z. decide whether to reject upp
SCORPION-xisa [38]
I dont know.....................
8 0
3 years ago
Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2) , and X(1, −2) . Trapezoid WKLX ​ is translated 4 units right and 3 uni
IRINA_888 [86]
<span>WKLX
W(2, −3),  
K(4, −3),
L(5, −2) ,
X(1, −2) 

TRANSLATED 4 UNITS RIGHT and 3 UNITS DOWN to produce W'K'L'X

4 units right means the x coordinate is affected. Since the moving to the right, we add 4 to the x values of each vertice.

W = 2 + 4 = 6
K = 4 + 4 = 8
L = 5 + 4 = 9
X = 1 + 4 = 5

3 units down means the y axis is affected. We add 3 to the value of y but keep the negative sign.
W = -3 + -3 = -6
K = -3 + -3 = -6
L = -2 + -3 = -5
X = -2 + -3 = -5

The correct answer is: </span><span>W′(6, −6), K′(8, −6), L′(9, −5) , and X′(5, −5)</span>
6 0
3 years ago
Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove
steposvetlana [31]

Answer:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}

Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

8 0
3 years ago
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