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Ira Lisetskai [31]
3 years ago
15

How can I do this I need help

Mathematics
2 answers:
Effectus [21]3 years ago
6 0

Answer:

V = 175

Step-by-step explanation:

Look at the attachment

never [62]3 years ago
5 0

Answer:

v = 175

Step-by-step explanation:

42 = 7 + v/5

35 = 7 + v/5

175 = v

You would subtract 7 on both sides to leave the V/5 on one side.

V over 5 means V is being divided by 5. So, doing the inverse operation would leave the V on it’s own.

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Tiles with numbers 1 through 9 are placed in a bag. What is the probability of choosing an even number that is greater than or e
Virty [35]

Answer:

3/9

Step-by-step explanation:

Okay, first let's list all the numbers between 1 and 9

1, 2, 3, 4, 5, 6, 7, 8, 9

Now let's list all of the even numbers from the previous list

2, 4, 6, 8

Now let's list all of the numbers from the previous list that are equal to, or greater than, 4

4, 6, 8

That's only three potential numbers, out of a total of nine numbers

So it's a 3/9 chance

7 0
2 years ago
3x+2y+2z = -2<br> 2x+y-z = -2<br> x-3y+z = 0
KatRina [158]

Here you go, I really hope this helps

If you can, it would be appreciated if you can mark this as Brainliest. I'm trying to rank up

7 0
3 years ago
Read 2 more answers
28 and/or 29 pls. Find the value of x and y.
djyliett [7]

Answer:

Step-by-step explanation:

for y

take 30 degree as reference angle using sin rule

sin 30=opposite/hypotenuse

1/2=9/y (do cross multiplication)

2*9=y

18=y

for x

using pythagoras theorem

a^2+b^2=c^2

9^2+x^2=18^2

81+x^2=324

x^2=324-81

x^2=243

x=\sqrt{243

x=9\sqrt{3

7 0
3 years ago
Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,
MrRissso [65]

Answer:

4.82\cdot 10^{-4}

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

n-1=51

Therefore, the probability of drawing a three at the 2nd draw is

p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784

Then, at the third draw, the previous 2 cards are not replaced, so there are now

n-2=50

cards in the deck. So, the probability of drawing a jack is

p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08

Therefore, the total probability of drawing an ace, a three and then a jack is:

p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}

4 0
4 years ago
If {4x-3y=17 and 2x-5y=-11 then y=?
anyanavicka [17]
You need to solve this "system of linear equations."  In other words, find a point (x,y) that satisfies both 4x-3y=17 and 2x-5y=-11.

Try solution by elimination.  Multiply the 2nd equation by -2 to obtain -4x+5y=22.  Add this result to the 1st equation.  I'd suggest you write this out to see what is happening.  

 4x-3y=17

-4x+10y=22
----------------
        7y=39.  Solving for y, we get y=39/7 (a rather awkward fraction).

Now find x.  To do this, substitute 39/7 for y in either of the given equations.  Solve the resulting equation for x.

Write your solution in the form (x, y):  ( ? , 39/7).
3 0
3 years ago
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