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3 years ago

Can someone help me with this math

Mathematics

1 answer:

astra-53 [7]3 years ago

6 0

Equal them to each other and you get that x=7

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Solve the inequality x + 3 < 7. Which number line represents the solution?<br> es )

Leya [2.2K]

Answer:

x < 4

So the dot on the graph would be an open dot and the number line would be shaded to the left.

Step-by-step explanation:

I do not see a number line, but

x < 4

So the dot on the graph would be an open dot and the number line would be shaded to the left.

6 0

3 years ago

I'm stuck on here, please, someone help me.

Sophie [7]

Answer:

Answer B

Step-by-step explanation:

therw is one full one and one with five

6 0

3 years ago

A wall is to be painted. A small rectangle in the middle of the wall will remain white to use for a projector. Find the area of

inna [77]

Answer:

108, 15, 93

Step-by-step explanation:

3 0

3 years ago

Solve for x: |2x + 12| =

igor_vitrenko [27]

Dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

6 0

3 years ago

Read 2 more answers

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

3 years ago