What is 2/5 x 15 and what is 7 x 3/7

Answers

Answer 1

Answer: The first one 2/6 x 15 is =6
The Second one 7 x 3/7 =3

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Square root of -25 in imaginary form

Answers

remember
√(mn)=(√m)(√n)

and
√-1=i

√-25=(√-1)(√25)=(i)(5)
aswer is +/-5i

root (-25) = root(-1) * root(25)
in imaginary form we know that root(-1) = i
so root(-25) = 5 i

There are a total of 30 white and black marbles in a box. There are 14 more white marbles than black marbles. How many black marbles are in the box?

Answers

Answer:

Step-by-step explanation:

30-14=16

Final answer:

The box contains 8 black marbles.

Explanation:

The question is asking about a box containing a total of 30 marbles, with the white marbles numbering 14 more than the black ones. So, we can first create an equation which equates the total number of marbles (30) to the sum of the black marbles and the white marbles. Let's denote the number of black marbles as B. The number of white marbles can then be expressed as B + 14.

Therefore, the equation would be: B + (B + 14) = 30, simplifies to 2B + 14 = 30.

To solve for B, subtract 14 from both sides of the equation so you have 2B = 16. Now, divide each side by 2 to solve for B, which gives us B=8.

Therefore, there are 8 black marbles in the box.

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Write an equation of a circle whose center is (-3,2) and whose diameter is 10

Answers

$The\ equation\ of\ a\ circle:(x-a)^2+(y-b)^2=r^2\n\n(a;\ b)-the\ coordinates\ of\ the\ cener;\ r-the\ radius\n-------------------------------\nThe\ center\ (-3;\ 2)\to a=-3;\ b=2\na\ diameter\ d=2r\ therefore\ r=(1)/(2)d\to r=(1)/(2)\cdot10=5\n\nAnswer:\n\boxed{(x-(-3))^2+(y-2)^2=5^2}\to\boxed{(x+3)^2+(y-2)^2=25}$

$(x-x_1)^2+(y-y_1)^2=r^2\n(x_1,y_1) - \text{ center}\n\nr=\frac{\text{ diameter}}{2}=(10)/(2)=5\n\n(x-(-3))^2+(y-2)^2=5^2\n\boxed{(x+3)^2+(y-2)^2=25}$

Elimination method 1) -4x + y = -12
4x + 2y = 6

2) 5x + 2y = 12
-6x - 2y = -14

3) 5x + 4y = 12
7x - 6y = 40

4) 5m + 2n = -8
4m + 3n = 2

Answers

3y=-6
y=-2
4x+2(-2)=6
4x-4=6
4x=10
x=5/2
(5/2,-2)

-x=-2
x=2
5(2)+2y=12
10+2y=12
2y=2
y=1
(2,1)

35x+28y=84
-35x+30y=-200
58y=-116
y=-2
5x+4(-2)=12
5x-8=12
5x=20
x=4
(4,-2)

20m+8n=-32
-20m-15n=-10
-7n=-42
n=6
5m+2(6)=-8
5m+12=-8
5m=-20
m=-4
(-4,6)

$1.)\n \n\begin{cases}-4x + y = -12\n 4x + 2y = 6 \end{cases}\n -------- \n 3y=-6/:3\n \ny=-2\n \n-4x+2=-12\n \n-4x=-12+2$

$-4x=-10/:(-4)\n \nx=(10)/(4)\n \nx=(5)/(2)\n \n\begin{cases}x=(5)/(2) \ny=-2\end{cases}$

$2.)\n \n\begin{cases}5x + 2y = 12\n -6x - 2y = -14 \end{cases}\n -------- \n -x=-2/*(-1)\n \nx=2\n \n5*2+2y=12$

$10+2y=12 \n2y=12-10\n \n2y=2/:2\n \ny=1\n \n\begin{cases}x=2 \ny=1\end{cases}$

$3.)\n \n\begin{cases} 5x + 4y = 12/*7\n 7x - 6y = 40/*(-5) \end{cases}\n \n\begin{cases} 35x + 28y = 84\n -35x +30y = -200 \end{cases}\n---------- \n58y=-116/:58\n \ny=-2$

$5x+4*(-2)=12\n \n5x-8=12\n \n5x=12+8\n \n5x=20/:5\n \nx=4\n \n\begin{cases}x=4\ny=-2\end{cases}$

$4)\n \n\begin{cases}5m + 2n = -8 /*4\n 4m + 3n = 2/(-5) \end{cases}\n \n \begin{cases}20m + 8n = -32\n -20m -15n = -10 \end{cases}\n------------ \n -7n=-42/:(-7)\n \nn=6$

$5m+2*6=-8\n5m+12=-8\n5m=-8-12\n5m=-20/:5\n \nm=-4\n \n\begin{cases}n=6\nm=-4\end{cases}$

How do you solve 5[2p−4(p+5)]=25 properly?