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2 years ago

Plzzz help plzzz I need help on both of them

Mathematics

1 answer:

Paul [167]2 years ago

5 0

The answer to the first problem is G.

Since there are 3 boys and 9 girls who are left-handed, that makes 12 people in total in sixth grade who are left-handed. The total number of sixth graders at the middle school is 150 since there are 90 girls and 60 boys in that grade. Divide 150 from 12 to get 0.08. You change it to percentage form by multiplying 0.08 by 100 to get 8%.

The answer for the second problem is C. You change the percentage (in this case it is 76%) to decimal form by dividing by 100 to get 0.76. You then multiply 0.76 with 8,950 to get 6,802.

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5(x+1)=20<br><br> What is is the answer?

Ilia_Sergeevich [38]

Steps to solve:

5(x + 1) = 20

~Distributive property

5x + 5 = 20

~Subtract 5 to both sides

5x + 5 - 5 = 20 - 5

~Simplify

5x = 15

~Divide 5 to both sides

5x/5 = 15/5

~Simplify

x = 3

Best of Luck!

3 0

3 years ago

Read 2 more answers

X ^ 2 + y ^ 2 + 6y - 67 = 2y - 6x; circumference

garri49 [273]

Answer:

(x+3)^2+(y+2)^2=80

Step-by-step explanation:

7 0

1 year ago

If (−4, 11) and (−6, 5) are the endpoints of a diameter of a circle, what is the equation of the circle?

lora16 [44]

Answer:

A. $(x+5)^2+(y-8)^2=100$

Step-by-step explanation:

1. The standard form of the equation of a circle is $(x-h)^{2} + (y-k)^2=r^2$

2. To find the center we require the midpoint of the 2 given points

center = $[\frac{1}{2}(-4-6),\frac{1}{2}(11+5)]$

=(-5,8)

3. The radius is the distance from the centre to either of the 2 given points calculate the radius using the distance formula

d= $\sqrt{(x_{2}+x_{1})^{2} +(y_{2}-y_{1})^{2}$

let $(x_{1},y_{1})= (-5,8) \\$ and $(x_2,y_2)=(-4,11)$

r= $\sqrt{(-4+5)^2+(11-8)^2} =\sqrt{81+9}=10$

--> $(x-(-5)^2+(y-8)^2=10^2$

--> $(x+5)^2+(y-8)^2=100$

4 0

2 years ago

Read 2 more answers

Properties in mathematics 0+14=14 is what type of property

sveta [45]

The zero propperty because you are add any thing plus zero

6 0

3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

2 years ago