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2 years ago

13

On the number line 0.2 would be located. Choose a answer to make a true statement.

1 answer:

Mariulka [41]2 years ago

6 0

Um, there are no choices, so I'll just tell you were it'd be located

Since 0.2 isn't yet 1, but its value is more than 0, it would be located in the middle of 1 and 0.

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I only need #16 this Test/homework is hard. Help

Amiraneli [1.4K]

Answer:

d

Step-by-step explanation:

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1 year ago

Complete the pairs of corresponding parts if △RST ≅ △TXY.

ELEN [110]

Answer:

∡S corresponds to ∡X

∡TXY = ∡RST

∡XTY = ∡SRT

∡XYT = ∡STR

Step-by-step explanation:

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2 years ago

Which pair of numbers has 2 as its greatest common factor

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Answer:

6 and 12 have the greatest common factor of 2

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2 years ago

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What is the point-slope form of a line with slope 3 that contains the point (2,1)

ryzh [129]

Answer:

Point slope form:

The equation of line is given by: $y-y_1=m(x-x_1)$ .....[1]

where m is the slope and a line contains a point $(x_1, y_1)$ .

As per the given statement:

Slope(m) = 3

$(x_1, y_1)$ . = (2, 1)

Substitute these in [1] we have;

$y-1=3(x-2)$

therefore, the point slope form of a line with slope 3 that contains the point (2, 1) is; $y-1=3(x-2)$

5 0

3 years ago

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Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

3 years ago