All categories

3 years ago

Can I get help plzzzzz

Mathematics

2 answers:

Alexandra [31]3 years ago

4 0

Answer:

no you cant

Step-by-step explanation:

Zina [86]3 years ago

4 0

Answer:

Please find attached pdf

Step-by-step explanation:

Download pdf

You might be interested in

If f(x)=-4x^2-6x-1 and g(x)=-x^2-5x+3, fine (f-g)(x)

NeX [460]

Answer:

$\large\boxed{(f-g)(x)=-3x^2-x-4}$

Step-by-step explanation:

$(f-g)(x)=f(x)-g(x)$

$f(x)=-4x^2-6x-1,\ g(x)=-x^2-5x+3\\\\(f-g)(x)=(-4x^2-6x-1)-(-x^2-5x+3)\\\\(f-g)(x)=-4x^2-6x-1-(-x^2)-(-5x)-3\\\\(f-g)(x)=-4x^2-6x-1+x^2+5x-3\qquad\text{combine like terms}\\\\(f-g)(x)=(-4x^2+x^2)+(-6x+5x)+(-1-3)\\\\(f-g)(x)=-3x^2-x-4$

5 0

3 years ago

A triangle with sides of lengths 16, 30, and 35 is a right triangle. O A. True O B. False

Afina-wow [57]

Answer:

B false

Step-by-step explanation:

The Theorem states that for a right triangle,

a^2 + b^2 = c^2

where c is the hypotenuse and a and b are the two shorter sides.

So does our given lengths work? Let's see:

16^2+30^2=35^2

256+900=1225

1156=1225

1156≠1225

And as we can see, the two sides do not equal each other, so this means the sides do not make a right triangle.

6 0

2 years ago

Read 2 more answers

-2(-2x + 3)-x -1 = -28

vekshin1

Answer:

x = -1

Step-by-step explanation:

I hope it helps u dear

5 0

2 years ago

Read 2 more answers

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

6 0

2 years ago

What is 32/23 as a decimal rounded to the nearest hundredth<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B32%7D%7B23%7D%20"

Softa [21]

Answer:

1.39

Step-by-step explanation

32/23 = 1.39130434783

round: 1.39

7 0

3 years ago