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3 years ago

Please help with the first question

Mathematics

1 answer:

NemiM [27]3 years ago

8 0

Answer:

<h2>I Think Letter A</h2>

Step-by-step explanation:

Hope It Helps

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Can someone help me please.

wolverine [178]

9514 1404 393

Answer:

$a=\sqrt{33}$

Step-by-step explanation:

The Pythagorean theorem tells you that the area of the square with side 'a' is the difference of the areas of the other two squares.

a² = 7² -4²

a² = 49 -16 = 33

a = √33

5 0

3 years ago

What is factor ? someone explain pls !!

Lubov Fominskaja [6]

Step-by-step explanation:

Factor, in mathmatechis, a number or algebric expression that divides another number or expression evenly.

For example, 3 and 6 are factors of 12 because 12 divided by 3 equals 4 exactly and 12 divided by 6 equals 2 exactly. The other factors of 12 are 1, 2, 4, and 12.

3 0

2 years ago

Read 2 more answers

Enter the simplified form of the complex fraction in the box. Assume no denominator equals zero.

Olenka [21]

Answer:

$\frac{(9-x)}{3x}$

Step-by-step explanation:

we have

$\frac{\frac{2}{x-3}-\frac{3}{x}}{\frac{3}{x-3}}$

step 1

Solve the numerator of the quotient

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

step 2

substitute in the original expression

$\frac{\frac{9-x}{(x-3)x}}{\frac{3}{x-3}}$

$\frac{(x-3)(9-x)}{3x(x-3)}$

step 3

simplify

$\frac{(9-x)}{3x}$

5 0

2 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:

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

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