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Yakvenalex [24]
3 years ago
9

Solve this system of linear equations. Separate

Mathematics
1 answer:
bearhunter [10]3 years ago
7 0

Answer:

(-9.6,2)

Step-by-step explanation:

hope i helped

pls can i get brainliest

-Zylynn

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What is the volume of a cube whose face has an area of 64cm^2?
MatroZZZ [7]

Answer:

512

Step-by-step explanation:

6 0
3 years ago
For f left parenthesis x right parenthesis equals StartRoot x EndRoot and g left parenthesis x right parenthesis equals 4 x plus
lara31 [8.8K]

Answer:

Step-by-step explanation:

Given that there are two functions f and g as

f(x) = \sqrt{x} \\g(x) =4x+9

We have to find the composition of functions.

Composition functions are calculated as the first function inside bracket and then the outside function of answer inside.

a)fog= f{g(x)} =f(4x+9) = \sqrt{4x+9}

b) gof = g{f(x)} = g(\sqrt{x} )=4\sqrt{x} +9

c) fof = f(\sqrt{x} ) = \sqrt[4]{x}

d) gog = g(4x+9) = 4(4x+9)+9\\= 16x+45

4 0
3 years ago
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

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\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

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1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

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)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

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\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

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6 0
4 years ago
Find the zeros of the function f(x) = x2 + 5x + 6. A) y = 6 because the graph crosses the y-axis at 6. B) y = -0.25 because that
Fofino [41]

Answer:

D.)

Step-by-step explanation:

The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts.  You can graph the function and see when the graph crosses the x-axis or solve for the x-values.  I will solve it via factoring and so:

f(x)=x^2+5x+6

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6.  Now let's think about all the factors of 6 we have: 6×1 and 2×3.  Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5?  Actually we can say 6-1=5 and 2+3=5.  Let's try both.

First let's use 6 and -1 and so:

x^2+5x+6\\\\x^2+6x-x+6\\\\x(x+6)-1(x-6)

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.

Now let's try 2 and 3 and so:

x^2+5x+6\\\\x^2+3x+2x+6\\\\x(x+3)+2(x+3)\\\\(x+2)(x+3)

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:

Case 1:

f(x)=x+2\\\\0=x+2\\\\x=-2

Case 2:

f(x)=x+3\\\\0=x+3\\\\x=-3

So your zero's are when x=-2 and x=-3.

D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.


~~~Brainliest Appreciated~~~

8 0
3 years ago
I need to factor this expression
soldier1979 [14.2K]

Answer:

( 5x + 1 ) ( x - 4 )

Step-by-step explanation:

5x² - 19x - 4

= 5x² + x - 20x - 4

= x ( 5x + 1 ) - 4 ( 5x + 1 )

= ( 5x + 1 ) ( x - 4 )

Hence, factorized.

6 0
2 years ago
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