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GenaCL600 [577]

3 years ago

5

Krya is playing a card game. Black cards are worth points while red cards take points away. What is her score if she has a black

20-card and a red 42-card

1 answer:

Nitella [24]3 years ago

8 0

Answer:

her score would be -22

Step-by-step explanation:

20-42=-22

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Step-by-step explanation:

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If matrix A has dimensions 3 x 2 and matrix B has dimensions 3 x 2, can matrix A and matrix B be multiplied?

Charra [1.4K]

Matrix A has 3 rows, 2 columns
Matrix B has 3 rows, 2 columns

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4 years ago

Read 2 more answers

I NEED HELP PLS THIS IS DUE IN 3 HOURS

Mariulka [41]

Answer:

Part 1) $x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})$

Part 2) $x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

Part 1)

in this problem we have

$x^{2} -2x-2=0$

so

$a=1\\b=-2\\c=-2$

substitute in the formula

$x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}$

therefore

$x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))$

$x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})$

Part 2)

in this problem we have

$x^{2} -6x+4=0$

so

$a=1\\b=-6\\c=4$

substitute in the formula

$x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(4)}} {2(1)}$

$x=\frac{6(+/-)\sqrt{20}} {2}$

$x=\frac{6(+/-)2\sqrt{5}} {2}$

$x_1=\frac{6(+)2\sqrt{5}}{2}=3+\sqrt{5}$

$x_2=\frac{6(-)2\sqrt{5}}{2}=3-\sqrt{5}$

therefore

$x^{2} -6x+4=(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))$

$x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$

5 0

3 years ago

Write the slope-intercept form of the equation of the line described.

iren2701 [21]

Answer:

9) y=3x+11

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Step-by-step explanation:

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6 0

3 years ago