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3 years ago

Any helpers?( it’s talking about the blue dot )

Mathematics

2 answers:

zheka24 [161]3 years ago

8 0

Answer:

Step-by-step explanation:

The answer would be no because the point is on the dotted line.

The inequality created has a dashed line which indicates that points on that line are not solutions to the inequality therefore the blue dot is not a solution.

scoray [572]3 years ago

4 0

Answer:

Step-by-step explanation:

The line is dotted, meaning that the inequality is y < 3 (y must be less than 3).

This means that the blue dot is not a solution because its coordinates are (2,3), which is not included.

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Too stoned pls help

Talja [164]

13x=65
65/13=5
So x equals 5

3 0

3 years ago

What is the value of a in the equation 5a -10b = 45 when b = 3

leva [86]

Salutations!

$5a-10b = 45$

value of $b = 3$

Multiply 10 and 3

$10*3=30$

Now, you can solve the equation --

$5a-30=45$

separate the variables aside, and the numbers aside.

$5a=45 +30$

$5a=75$

$a=75/5$

$a=15$

Hope I helped (:

Have a great day!

6 0

3 years ago

Pls help I need it now

Marina CMI [18]

Answer:

I think the answer is C sorry if I'm wrong but hope I helped

5 0

3 years ago

1 х Given g(x)=

yarga [219]

(a) Since $g(x)=\sqrt[3]{x}$ and $h(x) = \frac1{x^3}$ , we have

$(g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x$

We're given that

$(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}$

but we can rewrite this as

$\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}$

(bear in mind that we can only do this so long as x ≠ 0) so it follows that

$f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}$

(b) On its own, we may be tempted to conclude that the domain of $(f\circ g\circ h)(x) = \frac1{1+x}$ is simply x ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.

$g(x) = \sqrt[3]{x}$ is defined for all x - no issue here.

$h(x) = \frac1{x^3}$ is defined for all x ≠ 0. Then $(g\circ h)(x) = \frac1x$ also has a domain of x ≠ 0.

$f(x) = \frac1{1+x}$ is defined for all x ≠ -1, but

$(f\circ g\circ h)(x)=f\left(\frac1x\right) = \dfrac1{1+\frac1x}$

is undefined not only at x = -1, but also at x = 0. So the domain of $(f\circ g\circ h)(x)$ is

$\left\{x\in\mathbb R \mid x\neq-1 \text{ and }x\neq0\right\}$

7 0

3 years ago

Jack took 2 1/4 hours to complete a project. John took 3/8 hours longer to complete the same project. How long did John take to

Nutka1998 [239]

It takes Jack 2 1/4 hours (or 9/4 hours) to complete a project. John takes 3/8 hours longer.

We need to add 9/4 + 3/8

First, make the denominators equivalent.

We can do this by multiplying 2/2 to 7/2 which makes the denominator 8 (since 3/8 is already 8)

9/4 x 2/2 = 28/8
(note that we can do this because 2/2 is 1 and multiplying a number by 1 doesn't change its value)

Add 28/8 + 3/8
28/8 + 3/8 = 31/8

31/8 hours is our final answer, as a mixed number it is 3 7/8 hours :)

6 0

3 years ago