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bija089 [108]
3 years ago
13

Which is an equation of the line that passes through the points (5,2) and (10,-3)​

Mathematics
1 answer:
arsen [322]3 years ago
6 0

Answer:

y = x - 4

Step-by-step explanation:

equation of a line can be expressed as

(x - x1)/(x2 - x1) = (y - y1)/(y2 - y1)

(x - 5)/(10 - 5) = (y - 2)/(-3 - 2)

cross multiplying,

-5(x - 5) = 5(y - 2)

clearing the parentheses

-5x + 10 = 5y - 10

hence,

5y = 5x - 20

y = x - 4

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in which place should you write the first digit of the quotient of 3381 divided by 47 how can you determine that without using d
natima [27]
The 8 since u can't divide 3 or 33 by 47 so u have to divide by 338 first
5 0
3 years ago
The following graph shows the low temperatures for the past six nights. What was the range of temperatures for the six nights?
Kobotan [32]

Answer:

Step-by-step explanation:

highest temperature for past six nights = 34°F

lowest temperature for past six night = 22°F

so, the range of temperature = highest temperature - lowest temperature

                                                 = 34°F - 22°F

                                                 = 12°F

hence , the range of temperature is 12°F

3 0
4 years ago
Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.
rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed  as:

y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

Thus, the area of the surface is as follows:

\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA

= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]

3 0
3 years ago
And study 2.6 hours for the last exam. However there was a concert in town the night before and he scored 16 points lower than e
ollegr [7]

Answer: 67

Step-by-step explanation:

Here is the complete question:

A professor determined the relationship between the time spent studying (in hours) and performance on an exam.

Performance = 70.443 + 4.885 × (time)

Ann studied 2.6 hours for the last exam. However there was a concert in town the night before and her score was 16 points lower than expected. What was her score on this exam, rounded to the nearest integer?

Let's rewrite the formula to determine Ann's performance.

P = 70.443 + 4.885t

where t is in hours.

This is equation with P(t) means that P depends on variable t. We can then express t=2.6 in the formula to get her expected performance.

P = 70.443 + 4.885t

P = 70.443 + 4.885(2.6)

P = 70.443 + 12.701

P = 83.144

Now, since the question says that she scored 16 points less than the expected, we then need to find value of P-16

= P - 16

= 83.144 - 16

= 67.144

We can then round it to the nearest integer, this will be 67.

7 0
3 years ago
−4x+11y=15<br> ​x=2y<br> ​​
Vsevolod [243]
Hello

-4x + 11y = 15
x = 2y

Ok, we need to solve x = 2y for x
Lets start by substitute 2y for x in -4x + 11y = 15
-4x + 11y = 15
-4(2y) + 11y = 15
-8y + 11y = 15
3y = 15
Divide both sides by 3
3y/3 = 15/3
Y = 5

Again, substitute 5 for y in x= 2y
x = 2y
x = 2(5)
x = 10

Final answer : X = 10 and y = 5

Good luck!
3 0
4 years ago
Read 2 more answers
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