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2 years ago

7

The sum of two numbers is -18. If the first number is 10, which equation represents this situation, and what is the second numbe

r?

1 answer:

Oksana_A [137]2 years ago

6 0

Answer:

Step-by-step explanation:

8

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Help please this is due soon!!!!!!!

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

Slope of (-4,7),(-7,-15) is 22/3

Slope of the function is 2/11

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2 years ago

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Jim provides photos for two online sites: site A and site B. Site A pays $0.85 for every photo Jim provides. The amount in dolla

yarga [219]

We can use the same equation to figure out with only substituting the amount he was paid per photo.

Site A y = $0.85x = $0.85(5) = $4.25
Site B y = $0.40x = $0.40(5) = $2.00

We can see that Jim was paid $2.25 more at site A than Site B.

6 0

2 years ago

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. Given that O is the centre of the following circle, find the values of the unknowns.

alex41 [277]

Given:

Radius of circle is 13 cm

Height of triangle is 5 cm

To find:

Value of 'a' and 'b'

Steps:

To find value of 'a', we will use Pythagoras theorem as the triangle is a right angle triangle,

A² + B² = C²

$a^{2} + 5^{2} = 13^{2}$

$a^{2} + 25 = 169$

$a^{2} = 169 - 25$

$a^{2} = 144$

$\sqrt{a^{2}}=\sqrt{144}$

$a = 12$

Now to find the value of 'b', i will use law of cosine,

$c=\sqrt{a^{2}+b^{2}-2ab(cos\beta ) }$

$12 = \sqrt{13^{2}+5^{2}-2(5)(13)(cos\beta )}\\12=\sqrt{169 + 25-130(cos\beta )}\\12=\sqrt{194-130(cos\beta )}\\144 = 194 - 130(cos\beta )\\50 = 130(cos\beta )\\cos\beta = 0.3846\\\beta = cos^{-1}(0.3846)\\\beta = 67.38$

Therefore, the values of 'a' and 'b' is 12 and 67.38 respectively

Happy to help :)

If u need any help feel free to ask

3 0

2 years ago

Tom painted 3/5 of his bedroom on Monday and 1/10 of his bedroom on Tuesday. How much does he have left to finish?

guajiro [1.7K]

Answer:

3/10

Step-by-step explanation:

3/5 + 1/10

3/5 = 6/10

6/10 + 1/10 = 7/10 = total bedroom space he painted

10/10 - 7/10 = 3/10 = how much space is left

Answer = 3/10

5 0

3 years ago

Read 2 more answers

In this problem we will be dealing with the probability density function (pdf) associated with a continuous random variable, X.

konstantin123 [22]

Answer:

$b)\ 0.568$

Step-by-step explanation:

Given

$f(x) = 6x(1- x);\ 0 \le x \le 1$

Required

$P(0.3 < x < 0.7)$

From the question, we have:

$P(a < x < b) = \int\limits^b_a {f(x)} \, dx$

So, we have:

$P(0.3 < x < 0.7) = \int\limits^{0.7}_{0.3} {6x(1 - x)} \, dx$

Open bracket

$P(0.3 < x < 0.7) = \int\limits^{0.7}_{0.3} {6x - 6x^2} \, dx$

Integrate

$P(0.3 < x < 0.7) = \frac{6x^2}{2} - \frac{6x^3}{3}}|\limits^{0.7}_{0.3}$

$P(0.3 < x < 0.7) = 3x^2 - 2x^3|\limits^{0.7}_{0.3}$

Substitute 0.7 and 0.3 for x

$P(0.3 < x < 0.7) = (3*0.7^2 - 2*0.7^3) - (3*0.3^2 - 2*0.3^3)$

Using a calculator, we have:

$P(0.3 < x < 0.7) = (0.784) - (0.216)$

Remove brackets

$P(0.3 < x < 0.7) = 0.784 - 0.216$

$P(0.3 < x < 0.7) = 0.568$

3 0

2 years ago