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3 years ago

Find the slope of the line that passes through the points (0,-3) and (4, 5).

Mathematics

2 answers:

hoa [83]3 years ago

8 0

Answer: 2

Step-by-step explanation:

to find this you use the equation

y2-y1

--------

x2-x1

so with your points tuat would be

5+3/4-0 (it becomes a plus bc two negatives= a positive

8/4=2

nika2105 [10]3 years ago

5 0

The slope of the line is 2

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You are researching two glaciers.the first glacier flows at a rate of 75 meters in 3 days.the second glacier flows at a rate of

podryga [215]

Answer:

$1.3\ days$

Step-by-step explanation:

step 1

Find the units rate of the first glacier

$\frac{75}{3}=25\frac{meters}{day}$

using proportion

$\frac{25}{1}\frac{meters}{day}=\frac{200}{x}\frac{meters}{days}\\ \\x=200/25\\ \\x=8\ days$

step 2

Find the units rate of the second glacier

$\frac{107.5}{5}=21.5\frac{meters}{day}$

using proportion

$\frac{21.5}{1}\frac{meters}{day}=\frac{200}{x}\frac{meters}{days}\\ \\x=200/21.5\\ \\x=9.3\ days$

step 3

Find the difference

$9.3\ days-8\ days=1.3\ days$

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2 years ago

56 projects 5:9 ratio divide

Alex Ar [27]

The answer is 4:4 it takes 4 of each

4 0

2 years ago

“Two lines in a plane always intersect in exactly one point.”

arsen [322]

The answer is B.
Parallel lines as they are in the same plane but never meet.

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3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

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3 years ago

Point C is located at (1, 2), and point D is located at (-4,-2). Find the point that is the distance from point C

marishachu [46]

Answer:

The answer would be c

Step-by-step explanation:

i choosed c because u have da negative sign in front of 4 and 2 and positive numbers with 1 and 2

6 0

3 years ago

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