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Black_prince [1.1K]
2 years ago
9

Find the x and y intercepts of the equation. Express your answers as points. 5 x + 7 y = − 140

Mathematics
1 answer:
iVinArrow [24]2 years ago
5 0

Answer:

x-intercept(s): (-28, 0)

y-intercept(s): (0, -20)

Step-by-step explanation:

Try This Answer.

You might be interested in
0 ÷ 0 = 1
TEA [102]

Answer:

0/0 = undefined not 1

Step-by-step explanation: Undefined means there are no solutions possible for the operations in view. The example in question is itself wrong. n/0 is infinite but not undefined. 0/0 is undefined

7 0
2 years ago
(9+8)+6/3-7×2 how to slove
klemol [59]

Answer:

Step-by-step explanation:

There is a specific order to do these kind of problems..

first, we do everything in parenthesis...

(9 + 8) + 6/3 - 7 * 2

17 + 6/3 - 7 * 2

now, starting from the LEFT side, do all multiplication and division in the order it comes

17 + 2 - 14

now, starting from the LEFT side, do all addition and subtraction in the order it comes

19 - 14 = 4

so ur answer is 4

Learn this...order of operations....this is the order u do these problems in..

PEMDAS

P = parenthesis

E = exponents

M = multiplication

D = division

A = addition

S = subtraction

keep in mind, multiplication/division are done from left to right....u do not have to do multiplication before u do division....it just depends on which comes first when starting from the left side.....that also applies to addition/subtraction

5 0
3 years ago
Read 2 more answers
You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of
Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}

P_{1} is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is \frac{30}{50} = \frac{3}{5}.

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is \frac{33}{53}

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is \frac{36}{56}

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is \frac{39}{59}. So

P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588

P_{2} is the probability that we go R - R - B - R

P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788

P_{3} is the probability that we go R - B - R - R

P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076

P_{4} is the probability that we go R - B - B - R

P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511

P_{5} is the probability that we go B - R - R - R

P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733

P_{6} is the probability that we go B - R - B - R

P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493

P_{7} is the probability that we go B - B - R - R

P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476

P_{8} is the probability that we go B - B - B - R

P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419

So, the probability that this last marble is red is:

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

P = P_{1} + P_{2}

P_{1} is the probability that we go R-R-R-R. It is the same P_{1} from the previous item(the last marble being red). So P_{1} = 0.1588

P_{2} is the probability that we go B-B-B-B. It is almost the same as P_{8} in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490

P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078

There is a 20.78% probability that we actually drew the same marble all four times

3 0
3 years ago
Simplify<br><br> 21.75 <br><br> Thanks
tatyana61 [14]

Answer:

21 3/4

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Estimate the product 8.1 x 4.2
Korvikt [17]
Round 8.1 down to 8
Round 4.2 down to 4

8(4)= 32

The estimated product is 32.
Hope this helps!
6 0
3 years ago
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