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arsen [322]
2 years ago
10

I need the answer please

Mathematics
2 answers:
SVETLANKA909090 [29]2 years ago
8 0
You need to provide a picture of the prism...
SpyIntel [72]2 years ago
8 0
I cant really answer without the prism
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Literal Equations: 8c + 6j = 5p, for c
bixtya [17]
8c + 6j = 5p         Subtract 6j from both sides
       8c = 5p - 6j   Divide both sides by 8
         c = \frac{5p - 6j}{8}
3 0
3 years ago
For the matrices below. Find all (real) eigenvalues. <br><br> A= [7 9]<br> [0 9]
Jet001 [13]

Answer:

The answer is "9 and 7".

Step-by-step explanation:

Given:

A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]

Using formula:

|A-\lambda \cdot I|= 0\\\\

\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\

\to \lambda=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda=9\\\\

8 0
3 years ago
Given deductions for this week: federal tax $61.34, FICA $52.05, and state $7.92 what is the annual federal tax deduction?
stealth61 [152]

Given the weekly deductions raised, the annual Federal Tax deduction is $ 3,189.68.

Given that the deductions for the week were: Federal Tax $ 61.34, FICA $ 52.05, and State $ 7.92; To determine what is the annual Federal Tax deduction, the following calculation must be performed:

The weekly deduction must be multiplied by the number of weeks that a year has, to obtain the final amount of taxes.

  • $ 61.34 x 52 = X
  • $ 3,189.68 = X

Therefore, the annual Federal Tax deduction is $ 3,189.68.

Learn more in brainly.com/question/25225323

8 0
2 years ago
What is the sum of 2 3/4 and −2 3/4 ?
makvit [3.9K]

Answer:

2\frac{3}{4} + -2\frac{3}{4} = 0

Step-by-step explanation:

1st step convert into improper fraction:

===> 2\frac{3}{4} + -2\frac{3}{4}

===> \frac{16}{4} + -\frac{16}{4}

2nd step now adding the numbers:

( Plus sign and minus sign multiplies and hence the plus sign changes to minus)

===> \frac{16-16}{4}

Therefore the sum of 2\frac{3}{4} + -2\frac{3}{4} = 0

8 0
2 years ago
Find the following quotients, and write the quotient in standard form. (a)- x^2-9 ÷ x-3 (b)- x^3-27 ÷ x-3 (c) x^4- 81 ÷ x-3
IgorLugansk [536]

Answer:

(a) x+3

(b) x^2+3x+9

(c) (x+3)(x^2+9)

Step-by-step explanation:

We have given that

(a) \frac{x^2-9}{x-3}

From the algebraic identity we know that

a^2-b^2=(a+b)(a-b)

So \frac{x^2-9}{x-3}=\frac{(x+3)(x-3)}{x-3}=x+3

(b) \frac{x^3-27}{x-3}

We know the algebraic identity

a^3-b^3=(a-b)(a^2+ab+b^2)

So \frac{x^3-27}{x-3}=\frac{x^3-3^3}{x-3}=\frac{(x-3)(x^2+3x+9)}{x-3}=x^2+3x+9

(c) We have given \frac{x^4-81}{x-3}

We know the algebraic identity

a^2-b^2=(a+b)(a-b)

\frac{x^4-81}{x-3}=\frac{(x^2)^2-(3^2)^2}{x-3}=\frac{(x^2-9)(x^2+9)}{x-3}=\frac{(x+3)(x-3)(x^2+9)}{x-3}=(x+3)(x^2+9)

3 0
2 years ago
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