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3 years ago

PLEASE HURRY! 50 PIONT! IS TIMED TEST!

Mathematics

2 answers:

IceJOKER [234]3 years ago

8 0

Answer: 1, 2, and 4.

Unfortunately i'm not 100% sure, since I haven't taken this test while my friend has, but they told me they checked those answers and it got it right so its worth a shot. If you get it wrong though I deeply apologize.

leva [86]3 years ago

4 0

Answer:

Step-by-step explanation

30 square meters!

I really hope it's right

sorry if you get it wrong >m<

You might be interested in

The parking space shown at the right has an area of 209 ft2. A custom

jeyben [28]

Given:

The parking space shown at the right has an area of 209 ft².

A custom truck has rectangular dimensions of 13.5 ft by 8.5 ft.

To find:

Whether the truck fit in the parking space or not?

Solution:

If the area of the truck is more than the area of the parking space, then the truck cannot fit in the parking space otherwise it will fit in the parking space.

The area of a rectangle is:

$Area=length\times width$

Area of the rectangular truck is:

$Area=13.5\times 8.5$

$Area=114.75$

The area of the truck is 114.75 ft² which is less than the area of the parking space, 209 ft².

Therefore, the truck can fit in the parking space because the area of the truck is less than the area of the parking space.

3 0

3 years ago

Find the value of 15.0 NN in pounds. Use the conversions 1slug=14.59kg1slug=14.59kg and 1ft=0.3048m1ft=0.3048m.

finlep [7]

3.37 lb

Step-by-step explanation:

The question requires you to convert weight in Newtons to weight in pounds.

Given 15.0 N to convert to pounds, remember the conversion rate where;

1 Newton = 0.224809 pound-force

1 N= 0.224809 lb

15 N= ?

Perform cross-product

=15*0.224809

=3.37 lb

Learn More

Weight conversions:brainly.com/question/762013

Keywords : value,pounds,conversions,significant figures

#LearnwithBrainly

3 0

3 years ago

Match the parabolas represented by the equations with their foci.

Elenna [48]

Function 1 $f(x)=- x^{2} +4x+8$

First step: Finding when $f(x)$ is minimum/maximum
The function has a negative value $x^{2}$ hence the $f(x)$ has a maximum value which happens when $x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2$ . The foci of this parabola lies on $x=2$ .

Second step: Find the value of y-coordinate by substituting $x=2$ into $f(x)$ which give $y=- (2)^{2} +4(2)+8=12$

Third step: Find the distance of the foci from the y-coordinate
$y=- x^{2} +4x+8$ - Multiply all term by -1 to get a positive $x^{2}$
$-y= x^{2} -4x-8$ - then manipulate the constant of y to get a multiply of 4
$4(- \frac{1}{4})y= x^{2} -4x-8$
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is $12- \frac{1}{4}=11.75$

Hence the coordinate of the foci is (2, 11.75)

Function 2: $f(x)= 2x^{2}+16x+18$

The function has a positive $x^{2}$ so it has a minimum

First step - $x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4$
Second step - $y=2(-4)^{2}+16(-4)+18=-14$
Third step - Manipulating $f(x)$ to leave $x^{2}$ with constant of 1
$y=2 x^{2} +16x+18$ - Divide all terms by 2
$\frac{1}{2}y= x^{2} +8x+9$ - Manipulate the constant of y to get a multiply of 4
$4( \frac{1}{8}y= x^{2} +8x+9$

So the distance of focus from y-coordinate is $\frac{1}{8}$ to the north of $y=-14$
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: $f(x)=-2 x^{2} +5x+14$

First step: the function's maximum value happens when $x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25$
Second step: $y=-2(1.25)^{2}+5(1.25)+14=17.125$
Third step: Manipulating $f(x)$
$y=-2 x^{2} +5x+14$ - Divide all terms by -2
$-2y= x^{2} -2.5x-7$ - Manipulate coefficient of y to get a multiply of 4
$4(- \frac{1}{8})y= x^{2} -2.5x-7$
So the distance of the foci from the y-coordinate is - $\frac{1}{8}$ south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when $x=8.5$ and $y=79.25$ . The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when $x=-2.75$ and $y=-10.125$ . Manipulating coefficient of y, the distance of foci from y-coordinate is $\frac{1}{8}$ to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when $x=1.5$ and $y=9.5$ . The distance of the foci from the y-coordinate is $\frac{1}{8}$ to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

8 0

3 years ago

Describe the pattern.<br> 0.03, 0.3, 3, 30, 300, 3,000

kodGreya [7K]

The next number is 30,000. Each time you would multiply three.

Hoped this helped.

~Bob Ross®

5 0

3 years ago

Read 2 more answers

Checking the Mean Value Theorem:<br><br>Number 3

mrs_skeptik [129]

$\bf f(x)=x+\cfrac{1}{x}\qquad \left[\frac{1}{2},2 \right]\\\\
-----------------------------\\\\
\cfrac{df}{dx}=1+\left(-1x^{-2} \right)\implies \cfrac{df}{dx}=1-\cfrac{1}{x^2}
\\\\\\
f'(c)=1-\cfrac{1}{c^2}\quad \quad 1-\cfrac{1}{c^2}=\cfrac{f(2)-f\left( \frac{1}{2} \right)}{2-\frac{1}{2}}
\\\\\\
1-\cfrac{1}{c^2}=\cfrac{\frac{5}{2}-\frac{5}{2}}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=0
\\\\\\
1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm\sqrt{1}\implies c=\pm 1$

there's a quick graph below of the bounds and the tangent at "c"

not happening -2 or 2 will have a tangent parallel to a,b, needless to say -2 is out of the range [a,b] anyway, so the only value is really 1, on the positive 1st quadrant

6 0

3 years ago