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3 years ago

Find the value of x. And show work

Mathematics

1 answer:

il63 [147K]3 years ago

6 0

Answer

we know that all of these angles sum is 180 so therefore . . .

90 + 39 + 3x = 180

(90+39)=

129 + 3x = 180

(180-129) =

3x = 51

(51/3) =

x = 17

Hope this helps! Have a good day and remember to be tiger fierce!! :)

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Answer:

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Step-by-step explanation:

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3 years ago

Sharonda uses a blend of dark chocolate and milk chocolate to make the ice cream topping at her restaurant. She needs to buy 120

White raven [17]

Step-by-step explanation:

ATQ,

Let d be the number of kilograms of dark chocolate she buys and m be the number of kilograms of milk chocolate she buys.

She needs to buy 120 kg of chocolate in total for her next order, and her recipe calls for twice the amount of dark chocolate as milk chocolate.

So,

m = 2d .....(1)

m + d = 120 ...(2)

We can also solve the above equations,

Put m = 2d in equation (2)

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4 0

4 years ago

PLEASE HELP FAST

Pachacha [2.7K]

C, I was in class this morning and we say the formula for this is $6000 x 1.15 to the power of 5. That is $12068.14

Hope I could help :)

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4 years ago

Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum

lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

$\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\$

$\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\ (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\$

$=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3} \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\$

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3 years ago