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2 years ago

5

What is 6 1/8 as a decimal?

2 answers:

lora16 [44]2 years ago

5 0

Answer:

6.125

Step-by-step explanation:

1/8 = .125

zaharov [31]2 years ago

4 0

6.125 is equivalent to 6 1/8

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Viefleur [7K]

The slope m of a line passing through points P(a, b) and Q(c, d) is found using the formula:

$m= \frac{b-d}{a-c}$ .

Thus, the slope in A is $m= \frac{12-2}{-4-(-10)}=\frac{10}{6}=\frac{5}{3}$ .

The slope in B is $m= \frac{-4-10}{6-18}=\frac{-14}{-12}=\frac{7}{6}$ .

Now, to compare 7/6 to 5/3, we can write the second fraction as 10/6.

So, the slope in A is larger than the slope in B.

Answer: A

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3 years ago

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kifflom [539]

$\\ \sf\longmapsto \dfrac{x^2-1}{x^2+5x+4}\leqslant 0$

Solving denominator

$\\ \sf\longmapsto x^2+5x+4>0$

$\\ \sf\longmapsto x^2+4x+x+4>0$

$\\ \sf\longmapsto x(x+4)+1(x+4)>0$

$\\ \sf\longmapsto (x+4)(x+1)>0$

$\\ \sf\longmapsto x>-4\:or x>-1$

Hence x $\neq$ -1
x can't be 0 as it makes function undefined

$\\ \sf\longmapsto -4$

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2 years ago

Find the product of (x+3)2 <br><br> HURRY ILL GIVE YOU BRAINLIEST ANSWER

Lesechka [4]

The answer is 2x+6

Explanation: 2 times 3 is 6 and 2 times x is 2x

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3 years ago

Estimate 5,403 divided by 94

VLD [36.1K]

Answer:

60

Step-by-step explanation:

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3 years ago

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Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

7 0

3 years ago