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denis-greek [22]
2 years ago
7

Please help me, I will give brainliest! :D

Mathematics
1 answer:
solong [7]2 years ago
3 0

Answer:

Sorry i dont know but need the points

Step-by-step explanation:

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What's an excuse if you miss a class​
diamong [38]

Answer:

Say you ate something bad and you kept throwing up idrk

Step-by-step explanation:

5 0
3 years ago
Solve for x in the second equation and substitute it into the first equation. The result is .
Mice21 [21]
Where is the equation?
6 0
3 years ago
Simplify the expression (8x^3y^2/4a^2b^4)^-2
icang [17]
Simplify the following:
((8 x^3 y^2 a^2 b^4)/4)^(-2)
8/4 = (4×2)/4 = 2:
(2 x^3 y^2 a^2 b^4)^(-2)
Multiply each exponent in 2 x^3 y^2 a^2 b^4 by -2:
(x^(-2×3) y^(-2×2) a^(-2×2) b^(-2×4))/(2^2)
-2×3 = -6:
(x^(-6) y^(-2×2) a^(-2×2) b^(-2×4))/2^2
-2×2 = -4:
(y^(-4) a^(-2×2) b^(-2×4))/(2^2 x^6)
-2×2 = -4:
(a^(-4) b^(-2×4))/(2^2 x^6 y^4)
-2×4 = -8:
b^(-8)/(2^2 x^6 y^4 a^4)
2^2 = 4:
Answer: 1/(4 x^6 y^4 a^4 b^8) thus the answer is A
3 0
3 years ago
Write a mathematical sentence that expresses the information given below. Use t as your variable name. If necessary:Susanne chec
Feliz [49]

Answer:

t+8>82^{\circ}

Explanation:

We were given that:

Let "temperature" be represented by "t"

Suzan checks the temperature at 11:00. If the temperature rises by 8 more degrees, it will break the record high temperature for the day which is 82 degrees

This information is mathematically represented as:

t+8>82^{\circ}

7 0
1 year ago
To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI
nadya68 [22]

<u>Part 1) </u>To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}

in this problem we have

\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})

therefore

<u>the answer  Part 1) is</u>

Law of Sines

<u>Part 2) </u>To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

g^{2}=h^{2}+i^{2}-2*h*i*cos\ G

In this problem we have

g=HI

G=angle Beta

substitute

HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta

HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}

therefore

<u>the answer Part 2) is</u>

Law of Cosines

3 0
3 years ago
Read 2 more answers
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