please help ASAP! i have been struggling on this question and need help!

Answer:

  quotient term: 6

  dividend term: 102

Step-by-step explanation:

[tex]\dfrac{782}{17}=\dfrac{680+102}{17}=\dfrac{680}{17}+\dfrac{\boxed{102}}{17}=40 +\boxed{6} = 46[/tex]

Check the picture below.

The value of f(x) at x=4 is 2.

A function is an expression, rule, or law in mathematics that describes a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions are common in mathematics and are required for the formulation of physical relationships in the sciences.

We have to find the value of function for x=4.

This means that from the figure we have to find the value of y when x= 4.

Now, looking into the figure we have

f(4)= 2

f(0)= -2

f(-1)= -1

f(3)= 4

So, the required value is 2.

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Answer:

The sum of 8 and 1 is 9.

lol

Step-by-step explanation:

Answer:

1. <  2. < 3. > 4. = 5. >

Step-by-step explanation:

Answer:

<

Step-by-step explanation:

Good Question!

8/20=4/10=2/5=0.4

7/13= lcm of 5 and 13 is 75.

26/75 [<] 35/75

:)

Answer: < less than

Step-by-step explanation:

1. Find LCM of 20 and 13.

20: 20,40,60,80,100,120,140,160,180,200,220,240,260,280

13: 13, 26, 39, 52, 65, 78, 91, 104, 107, 110 ...... 234, 247, 260

The LCM is 260.

2. 20*13 = 260 13*20=260

therefore we need to multiply the numerator by those 13 and 20.

8*13 = 104 7*20 = 140

3. compare Fractions:

(104/260) < (140/260)

[tex]x = 23[/tex]

Step-by-step explanation:

First, we need to write the equations in their standard form, i.e., all the variables and their coefficients are to be placed on the left hand side and the plain numbers on the right hand side. The first equation is already in its standard form:

[tex]2x + 5y = 35\;\;\;\;\;(1)[/tex]

but we need to rewrite the 2nd one. So the standard form of the 2nd equation can be written as

[tex]x + 15y = -10\;\;\;\;\;(2)[/tex]

Now that the equations are in their standard forms, the next step that we need to do is to eliminate one of the variables, namely the y. To do that, we need to multiply either one of the equations so that when we add them together, the y-variable gets eliminated. Note if we multiply Eqn(1) by -3, we'll get -15y on one equation and a +15y on the other equation so that when they are added together, they cancel out. So if we do that, we get

[tex]-6x - 15y = -105\;\;\;\;\;(3)[/tex]

Adding Eqn(3) to Eqn(2), we get

[tex]-5x = -115[/tex]

or

[tex]x = \dfrac{-115}{-5} = 23[/tex]

Answer:

I think this is what is wanted

Step-by-step explanation:

Answer:

$1650 is the answer hope it helps

Answer:

1231.50432021 cm³

Step-by-step explanation:

Use the formula:

[tex]\frac{1}{3} \pi r^{2} h[/tex]

And replace known values:

[tex]\frac{1}{3} \pi (7^{2})(24)[/tex]

Aaaand simplify:

[tex]\frac{1}{3}*1176 \pi[/tex]

Which gives approximately:

[tex]1231.50432021[/tex]

And don't forget to put the unit (cm³)!

Answer:

1230

Step-by-step explanation:

1/3*pi*7^2*24

1/3*pi*49*24

=1230

Answer:

-1.5

Step-by-step explanation:

Answer:

16x7 = 112 (16 ounces, 7 hours) 112 ounces to pints = 7 pints of water

Step-by-step explanation:

Answer: 7/12 miles

Step-by-step explanation:

20mph = 20/60 miles per minute = 1/3 miles per minute

1 min 45s = 1 min + 3/4 min = 7/4 min

1/3 x 7/4 = 7/12 miles

The distance travelled by car with the given speed and time is 7/12 miles.

The speed formula can be defined as the rate at which an object covers some distance. Speed can be measured as the distance travelled by a body in a given period of time. The SI unit of speed is m/s.

Given that, a car is traveling at 20 miles per hour during rush hour.

Time =1 minute and 45 seconds

= [tex]1\frac{45}{60}=1\frac{3}{4}[/tex]

= 7/4 minutes

Now, 20 miles per minute

= 20/60 miles per minute = 1/3 miles per minute

We know that, Distance =Speed×Time

= 1/3 ×7/4

= 7/12 miles

Therefore, distance travelled by car is 7/12 miles.

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Answer:

The answer is 4

Step-by-step explanation:

(-2.4)(10) / -3(2)

-24/-6

=4

Answer:

The answer will be 4! Hope this helps

Answer:

[tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

Step-by-step explanation:

I'm assuming the system is [tex]\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.[/tex]:

[tex]x^2+y^2=2[/tex]

[tex]x^2+(2x^2-3)^2=2[/tex]

[tex]x^2+(4x^4-12x^2+9)=2[/tex]

[tex]x^2+4x^4-12x^2+9=2[/tex]

[tex]4x^4-11x^2+9=2[/tex]

[tex]4x^4-11x^2+7=0[/tex]

[tex]x^4-11x^2+28=0[/tex]

[tex](x^2-7)(x^2-4)=0[/tex]

[tex](4x^2-7)(x^2-1)=0[/tex]

[tex]4x^2-7=0[/tex]

[tex]4x^2=7[/tex]

[tex]x^2=\frac{7}{4}[/tex]

[tex]x=\pm\sqrt{\frac{7}{4}}[/tex]

[tex]x^2-1=0[/tex]

[tex]x^2=1[/tex]

[tex]x=\pm1[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm\sqrt{\frac{7}{4}})^2-3[/tex]

[tex]y=2({\frac{7}{4}})-3[/tex]

[tex]y=\frac{7}{2}-3[/tex]

[tex]y=\frac{1}{2}[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm1)^2-3[/tex]

[tex]y=2(1)-3[/tex]

[tex]y=2-3[/tex]

[tex]y=-1[/tex]

Therefore, [tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

[tex]\dfrac 25 (d-10) - \dfrac 23(d+6)\\\\\\=\dfrac{6}{15} (d-10) - \dfrac{10}{15} (d+6)\\\\\\=\dfrac{6(d-10) -10(d+6)}{15}\\\\\\=\dfrac{6d -60-10d-60}{15}\\\\\\=\dfrac{-4d-120}{15}\\\\\\=-\dfrac{4d}{15} - 8[/tex]

Answer:

1.2

Step-by-step explanation:

First, you cancel out [tex]\frac{6}{20}[/tex] to become [tex]\frac{3}{10}[/tex]

Then that becomes [tex]-\frac{3}{10}+1.5[/tex]

Now, you convert the element to a decimal form to be [tex]-0.3+1.5[/tex]

Finally, you subtract them (because a negative plus a positive is subtraction) to get the final answer of 1.2

Hope this helps!

The domain and the range of the function describes the possible profit or

loss from the sale of popcorn and the number of popcorn units.

The correct responses are;

Reasons:

The question parameters are;

The price of a popcorn = $1.5

Number of popcorn he purchases = 150

Total profit made = $5.50

The number of containers sold = 25

Therefore;

C. Given that the number of containers sole is countable, the function will be discrete.

D. Cost of the containers = Revenue - Profit

Which gives;

Cost of the containers = 1.5 × 25 - 5.5 = 32

The given function is therefore;

y = 1.5·x - 32

E. The domain is the 150 containers which gives;

The domain = 0 ≤ x ≤ 150

F. When no popcorn is sold, we have;

x = 0, which gives;

y = 1.5 × 0 - 32 = -32

If he does not sell any popcorn, the loss is $32

G. If all the popcorn are sold, we have;

x = 150

Therefore;

y = 1.5 × 150 - 32 = 193

The profit if all the popcorn are sold = $193

H. The range of the function are the possible y values, which is the possible profit or loss, which gives;

The range = -32 ≤ y ≤ 193

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[tex]\textit{volume of a cylinder}\\\\ V=\pi r^2 h~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{648\pi }\\ h=8 \end{cases}\implies 648\pi =\pi r^2(8) \\\\\\ \cfrac{648\pi }{8\pi }=r^2\implies 81=r^2\implies \sqrt{81}=r\implies \boxed{9=r}[/tex]

Answer:

What's the question lol

[tex]y = x^{\tfrac 1{\ln x}}\\\\\implies \ln y = \ln \left( x^{\tfrac 1{\ln x}} \right)\\\\\implies \ln y = \dfrac{1}{\ln x} \cdot \ln x\\\\\implies \ln y = 1\\\\\implies y =e\\\\\\\text{Now,}\\\\\ \lim \limits_{x \to \infty} x^{\tfrac 1{\ln x}}=\lim \limits_{x \to \infty} y = \lim \limits_{x \to \infty} e = e[/tex]

The values of ||-10v|| is 51.96 units

Given the complex number v = 6i + 3j,

10v is expressed as 10(6i + 3j)

10(6i + 3j) = 60i + 30j

Take the modulus of the complex number:

[tex]|10v|=\sqrt{60^2+30^2}\\|10v|= \sqrt{1800+900}\\|10v|= \sqrt{2700} \\|10v|=51.96units[/tex]

Hence the values of ||-10v|| is 51.96 units

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Answer:

Yes.

Step-by-step explanation:

Associative property of multiplication is:

(A×B)×C = A×(B×C).

It says that "The way of grouping of terms doesn't change the product."

So, given example (2×3)×4=2×(3×4) is an example of associative property of multiplication.

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

[tex]\begin{cases} x = -4\implies &x+4=0\\ x=-2\implies &x+2=0\\ x=1\implies &x-1=0 \end{cases}\qquad \implies (x+4)(x+2)(x-1)=\stackrel{y}{0}[/tex]

now, that's the equation or polynomial in factored form, hmmm we also know that it has a y-intercept of -11, namely, when x = 0  y = -11, well let's plug in a factor to it, that will reflect those values, namely say hmmm factor "a", so

[tex](x+4)(x+2)(x-1)=y\qquad \stackrel{\textit{adding "a" factor for vertical shift}}{a(x+4)(x+2)(x-1)}=y \\\\\\ \stackrel{\textit{we know that when x = 0, y = -11}}{a(0+4)(0+2)(0-1)=-11}\implies -8a=-11\implies a=\cfrac{-11}{-8}\implies a = \cfrac{11}{8} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\mathbb{FOIL}}{\cfrac{11}{8}(x^2+6x+8)}(x-1)=y\implies \cfrac{11}{8}(x^3+6x^2+8x-x^2-6x-8)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{11}{8}(x^3+5x^2+2x-8)=y~\hfill[/tex]

Answer:

2

Ok lets start off with your 3 numbers

First do 10 / 5 = 2

Then do 12 / 6 = 2

After that do 20 / 10 = 2

now you see you that its all equivalent to 2 that's your answer!