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2 years ago

How can I solve this?? Please help

Mathematics

1 answer:

OLga [1]2 years ago

8 0

Answer:yes

15x times 2=30x

15x times 3= 45x

3y tines 2 =6y

3y times 3=9y

Step-by-step explanation:

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Not sure if I did this correctly... please help 3. Is the one I'm kinda stuck on

denis23 [38]

You got the b part wrong. It would be 4B+4C/2=4D

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3 years ago

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10^x • 100^2x = 1000^5. Step by Step

nordsb [41]

Step one: set the base of each term to be the same (10)
(10ˣ)(10²)²ˣ = (10³)⁵
(10ˣ) (10⁴ˣ)= (10¹⁵)
10⁵ˣ = 10¹⁵
Because the indices are the same we can equate the two.
5x = 15
x = 3

Check 10⁵ˣ³ = 10¹⁵ 10¹⁵=10¹⁵

4 0

3 years ago

Suppose that we have two identical boxes: box 1 and box 2. box 1 contains 5 red balls and 3 blue balls. box 2 contains 2 red bal

sineoko [7]

(a) The probability of drawing a blue marble at random from a given box is the number of blue marbles divided by the total number of marbles. We assume that the probability of selecting one of two boxes at random is 1/2 for each box.

... P(blue) = P(blue | box1)·P(box1) + P(blue | box2)·P(box2) = (3/8)·(1/2) + (4/6)·(1/2)

... P(blue) = 25/48 . . . . probability the ball is blue

(b) P(box1 | blue) = P(blue & box1)/P(blue) = (P(blue | box1)·P(box1)/P(blue)

... = ((3/8)·(1/2))/(25/48)

... P(box1 | blue) = 9/25 . . . . probability a blue ball is from box 1

8 0

3 years ago

Hey can anyone pls answer these math problems!!

Zepler [3.9K]

Answer:

1={5 and 3} 2={0} 3={0.209 and 4.791} 4={-3 and 4} 5={1 and 9} 6={-4.303 and -0.697}

Step-by-step explanation:

Your probably confused on what a, b, and c means.

The problems are written like this.

$ax^2+bx+c=0$

That formula finds the zeros.

For example 1:

$x^2-8x+15=0\\ax^2-bx-c\\a=1\\b=-8\\c=15$

Then use the formula

$x=\frac{-b+-\sqrt{b^2-4ac} }{2a} \\\\x=\frac{-(-8)+-\sqrt{-8^2-4(1)(15)} }{2} \\x=\frac{8+-\sqrt{64-60} }{2} \\\\x=\frac{8+-2}{2}\\$

If plus then 5, if minus then 3

CHECK:

$x^2-8x+15=0\\5^2-8(5)+15=0\\25-40+15=0\\0=0\\3^2-8(3)+15=0\\9-24+15=0\\0=0$

Hope this helps you understand how to do this by yourself. Now to do the rest ;-;

5 0

2 years ago

Determine which equation is parallel to the line JK and which is perpendicular to like JK

sattari [20]

Answer:

If the equation of JK is y = mx + c, then y = mx + c' is the equation of the straight line parallel to above and $y = - \frac{1}{m}x + b$ will be perpendicular to that.

Step-by-step explanation:

Let us assume that the line JK has the equation in slope-intercept form as

y = mx + c ............. (1)

Therefore, the equation has slope = m

Now, any straight line having an equation with slope m will be parallel to line JK.

So, the equation of the straight line which is parallel to equation (1) will be

y = mx + c', where, c' is any real constant.

Now, let us assume that another straight line having equation

y = nx + a is perpendicular to the line JK i.e. equation (1).

Now, we know if two lines are perpendicular to each other then the product of their slopes will be - 1.

So, mn = - 1

⇒ $n = - \frac{1}{m}$

Therefore, the equation of a straight line which is perpendicular to equation (1) will be $y = - \frac{1}{m}x + b$ where b is any real constant. (Answer)

3 0

3 years ago

Read 2 more answers