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I am Lyosha [343]

1 year ago

5

There are 25 students in Ms. Nguyen's second-grade class. In the class election, 4 students voted for Benjamin, 12 voted for Sah

il, and 999 voted for Maria.
What percentage of the class voted for Maria?

1 answer:

Yanka [14]1 year ago

7 0

The answer is 36% of students voted for maria

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ABCD is a parallelogram.<br>Angle CDA is 56°<br>What is the measure of angle ABC?

BartSMP [9]

Answer:

Step-by-step explanation:

angle CDA = angle ABC (opposite angles of a parallelogram are equal)

56 degree = angle ABC

therefore angle ABC is 56 degree.

6 0

3 years ago

Find the difference quotient and simplify your answer. f(x) = 7x − x2, f(7 + h) − f(7) h , h ≠ 0

uranmaximum [27]

Whatever is inside of the ( ), simply plug that digit into the x-values for f(x)
So: f(x) = 7x - x^2, and f(7+h) - f(7)
= [7(7+h) - (7+h)^2] - [7(7) - (7)^2]
= [49+7h - 49+14h+h^2] - [49-49]
= 49-49 + 7h+14h + h^2 = h^2 + 21h =
h (h+21), h (h+21) = 0
h=0... But it stated h cannot = 0
So h+21 = 0, h = -21

4 0

2 years ago

A triangle has the sides of 7 in, 14 in, and 25 in. Is it a right triangle?

MariettaO [177]

Answer:

No it does not equal a right triangle.

3 0

2 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

<span>The standard form of the equation of a circumference is given by the following expression:

</span> $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$
<span>
On the other hand, the general form is given as follows:

</span> $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ <span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span> $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

<em><u>First. a) matches 5) </u></em>

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

<em><u>Second. b) matches 1) </u></em>

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

<em><u>Third. c) matches 3)</u></em>

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

<em><u>Fourth. d) matches 6)</u></em>

6 0

3 years ago

Read 2 more answers

This is the problem that I need help with please. :)

Anna35 [415]

Answer:

umm I think it 186

Step-by-step explanation:

8 0

3 years ago