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3 years ago

13

jake reads for 20 minutes after school. he reads x more minutes after dinner. enter an equation that represents the total number

of minutes, y, that jake read

1 answer:

Nesterboy [21]3 years ago

4 0

Answer: 20 + x

Why you ask? Well its just plain logic if you read it thoroughly

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Solve the equation. (enter your answers as a comma-separated list.)2^−125x = (0.5)^x − 5

ivolga24 [154]

2 - 125x = 0,5x - 5
-125x - 0,5x = -5 -2
-125,5x=-7
x= -7/-125,5
x=0,05...

3 0

3 years ago

This morning, some dragons, horses, and chickens were playing in my backyard. I counted 15 heads, 50 legs, and 4 wings. How many

klasskru [66]

We are given

number of heads =15

we know that

any healthy dragon has three heads

horse has 1 head

chicken has 1 head

Let's assume

number of dragons is x

number of horses is y

number of chickens is z

so, we will get

first equation:

$3x+y+z=15$

number of legs =50

any healthy dragon has four legs

chicken has 2 legs

horse has four legs

so, we can get second equation as

$4x+4y+2z=50$

we can simplify it

$2(2x+2y+z)=50$

$2x+2y+z=25$

now, we can find third equation

dragon has two wings

horse has no wings

chicken has two wings

so, we will get third equations as

$2x+0y+2z=4$

now, we can simplify it

$2x+2z=4$

$2(x+z)=4$

$x+z=2$

so, we will get system of equations as

$3x+y+z=15$

$2x+2y+z=25$

$x+z=2$

now, we can use substitution

We can find for z from third equation

$z=2-x$

we can plug this in first equation

$3x+y+2-x=15$

now, we can solve for y

$2x+y+2=15$

$y=13-2x$

now, we can plug this z and y into second equation

$2x+2(13-2x)+2-x=25$

now, we can solve for x

$x-4x+26=23$

$-3x=-3$

$x=1$

now, we can find y and z

$y=13-2x$

we can plug x=1

$y=13-2*1$

$y=11$

$z=2-x$

we can plug x=1

$z=2-1$

$z=1$

Hence ,

number of dragons is 1

number of horses is 11

number of chicken is 1............Answer

7 0

3 years ago

Number 11. Thank you...

jekas [21]

The answer is B : 720
If you want to learn more on HOW to do this , message me :) !

5 0

4 years ago

I need help for b) so Please help

murzikaleks [220]

Answer:

5

Step-by-step explanation:

There are 5 domains

3 0

3 years ago

Evaluate the following when $a = 3$, $b = \frac14$, and $c = 5$: \[ \frac{b^{-1}}{ca^{-1} - ac^{-1}}. \] Thank you so much!

pickupchik [31]

$\dfrac{b^{-1}}{ca^{-1}-ac^{-1}}=\dfrac{\frac1b}{\frac ca-\frac ac}$

Multiply the numerator and denominator by $ac$ :

$\dfrac{\frac1b}{\frac ca-\frac ac}\cdot\dfrac{ac}{ac}=\dfrac{\frac{ac}b}{c-a}$

Since $b=\frac14$ , we have $b^{-1}=\frac1b=4$ . So substituting the values of <em>a</em>, <em>b</em>, and <em>c</em> gives us

$\dfrac{\frac{ac}b}{c-a}=\dfrac{3\cdot5\cdot4}{5-3}=\dfrac{60}2=\boxed{30}$

4 0

3 years ago

Read 2 more answers