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3 years ago

Solve for y. 15x -5y = 10thank you.

Mathematics

2 answers:

Setler79 [48]3 years ago

6 0

Answer:

Step-by-step explanation:

15x - 10 = 20

15x = 20 + 10

15x = 30 (divide both sides by 15 to get x)

15x/15 = 30/15

x = 2

Molodets [167]3 years ago

3 0

Answer:

y & x = 2, but since you only need y, y=2

Step-by-step explanation:

15 x 2 = 30

5 x 2 = 10

30 - 10 = 20

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You have 6 pints of glaze. It takes 7/8 of a pint to glaze a bowl and 9/16 of a pint to glaze a plate.

son4ous [18]

Answer:

a. How many bowls could you glaze?

6 bowls

How many plates could you glaze?

10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

How many plates can you glaze?

2 plates

How much glaze will be left over?

8/9 of glaze would be left

c. How many of each object could you glaze so that there is no glaze left over? Explain how you found your answer.

4 4/23 bowls and 4 4/23 plates

Step-by-step explanation:

We are told in the question

Number of pints of glaze = 6 pints

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

How many bowls could you glaze?

7/8 pint = 1 bowl

6 pints = x

Cross Multiply

x = 6 pints/ 7/8

= 6 × 8/7

= 48/7

= 6.8571428571 bowls

We can only glaze whole numbers of a bowl, hence, we can only glaze 6 bowls

How many plates could you glaze?

9/16 pint = 1 bowl

6 pints = y

y = 6 pints/ 9/16

= 6 × 16/9

= 10.666666667 plates

We can only glaze whole number of a plate, hence, we can only glaze 10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

1 bowl = 7/8 pints

5 bowls = z

Cross Multiply

z = 5 × 7/8pints

= 4 3/8 pints of glaze would be used for bowls

How many plates can you glaze?

The rest is for plates, hence:

The amount of glaze left for plates is calculated as:

6 pints - 4 3/8

1 5/8 pints of glaze would be left over to glaze plates.

So therefore,

9/16 pints = 1 plate

1 5/8 pints =

= 2 8/9

Hence we can only glaze 2 plates

How much glaze will be left over?

2 8/9 pints - 2 pints

= 8/9 pints of glaze.

c. How many of each object could you glaze so that there is no glaze left over?

We have 6 pints of glaze

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

Let the number of each objects be represented by x

7/8 × x + 9/16 × x = 6 pints

= 4 4/23 bowls and 4 4/23 plates

3 0

3 years ago

14.914 cm + 243.1 cm + 243.1 cm =269.0186 cm

Alexeev081 [22]

Answer:

False

Step-by-step explanation:

3 0

3 years ago

If you select a card at random, what is the probability of getting:A number smaller than 5 (counting the ace as a 1)?

garri49 [273]

4 Aces, 4 2's, 4 3's 4 4's

= 16 cards total out of 52 cards

so probability would be 16/52 which reduces to 4/13

4 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Daniel drew a toy from a treasure chest containing 30 different toys. He selected a bouncy ball without replacement and then sel

borishaifa [10]

Answer:

$Pr = \frac{1}{870}$

Step-by-step explanation:

Given

$n =30$

Required

Probability of selecting 2 toys of different types

From the question, we understand that all toys are different i.e. 1 of each type.

And the selection is without replacement;

So, after the first toy is selected; there are n - 1 toys left.

So, the probability is:

$Pr = \frac{1}{n} * \frac{1}{n - 1}$

Substitute $n =30$

$Pr = \frac{1}{30} * \frac{1}{30- 1}$

$Pr = \frac{1}{30} * \frac{1}{29}$

$Pr = \frac{1}{30*29}$

$Pr = \frac{1}{870}$

5 0

3 years ago

Solve for y. 15x -5y = 10thank you.​

Solve for y. 15x -5y = 10thank you.