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2 years ago

Help me please, i need an answer asap

Mathematics

1 answer:

Artyom0805 [142]2 years ago

7 0

Answer:

answer is A - library books by genre and by Adult or Child

Step-by-step explanation:

your column on the left will have different genre of books

your top column will have children or adults

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Iestion 10 (5 points)

marta [7]

Answer:

Jake is 9, and Jake's dad is 43.

Step-by-step explanation:

Ok, let's start off by putting a variable for Jack's age, A.

We know that Jack's father is 2 less than 5 times, so we can write his as 5x -2.

Now let's make an equation:

52 = x + 5x - 2

We need to isolate the x-values so we need to add 2 on both sides:

54= x + 5x

No we need to add the like terms:

54 = 6x

Finally, we just divide by 6 on each side.

Now we know X = 9 but we're not done yet :)

So because Jake's dad is almost 5 times as much, we first multiply Jack's age by 5. We get 45 but we need to subtract 2 which leaves us with 43.

3 0

3 years ago

Given that cos θ = –12∕13 and θ is in quadrant III, find csc θ.

Trava [24]

Answer:

Step-by-step explanation:

We are given:

$\displaystyle \cos(\theta)=-\frac{12}{13},\,\theta\in\text{QIII}$

Since cosine is the ratio of the adjacent side over the hypotenuse, this means that the opposite side is (we can ignore negatives for now):

$o=\sqrt{13^2-12^2}=\sqrt{25}=5$

So, the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.

And since θ is in QIII, sine/cosecant is negative, cosine/secant is negative, and tangent/cotangent is positive.

Cosecant is given by the hypotenuse over the opposite side. Thus:

$\displaystyle \csc(\theta)=\frac{13}{5}$

Since θ is in QIII, cosecant must be negative:

$\displaystyle \csc(\theta)=-\frac{13}{5}$

Our answer is A.

7 0

3 years ago

Read 2 more answers

Describe the slope of the line<br> Find the slope

Dmitry_Shevchenko [17]

Answer:

Question one: Zero slope
Question two: $m=\frac{0}{4}$

Step-by-step explanation:

Given the following questions:

<u>Question one:

</u>The following line is what you call a "zero slope." Zero slopes are lines that are neither decreasing or increasing and remain at a constant or just a straight line.

Question two:

Point A = (-2, -3) = (x1, y1)
Point B = (2, -3) = (x2, y2)

Using the formula for slope or rise over run we will solve and find the slope of this line.

$m=\frac{y2-y1}{x2-x1}$
$m=\frac{-3--3}{2--2} =\frac{0}{4}$
$m=\frac{0}{4}$

The slope of this line is "0/4."

Hope this helps.