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3 years ago

Please help!!! How do you count (-3y^2+5y) (4y+7)??

Mathematics

1 answer:

levacccp [35]3 years ago

6 0

You must distribute each value like this:

$(-3y^2+5y) (4y+7)\\ \\ (-12y^3-21y^2+20y^2+35y)\\ \\ (-12y^3-y^2+35y)$

Hope that helps!

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The points plotted below satisfy a polynomial equation. What type of extreme value does the graph have in the part that is visib

bezimeni [28]

Hi!

The graph shows an A) Maximum

A Maximum value appears in a graph when all other values of the polynomial are smaller in value i.e. are under it in an X-Y plot. This is expressed mathematically as:

There is a maximum if for a given x*: f(x*) ≥ f(x) for all x.

In the graph, you can clearly see that there is a value that is higher than all the others, so this value is a Maximum.

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3 years ago

What’s the domain and range of y=x^2+2

denpristay [2]

Answer:

domain: all real numbers

range: {y element R : y>=2}

Step-by-step explanation:

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3 years ago

Which number has two lines of symmetry? 101 619 33 18

7nadin3 [17]

33 in middle - and in between I

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3 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

The point-slope of a line is:

$y-y_1=m(x-x_1)$

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

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

3 years ago

The irrational number square root 67 is greater than [blank]

Tanya [424]

Answer:

The irrational number square root 67 is greater than whole number 8

Step-by-step explanation:

Given

$\sqrt{67$

Required

Complete the sentence

No options are given.

So, we will solve the expression then compare it with the nearest smallest integer

First, we solve the given expression

$\sqrt{67$

$\sqrt{67} \approx 8.185$

By comparison:

$8.185 > 8$

This implies that, the blank will be completed with 8

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3 years ago