Translate into algebraic terms...7 less than 5 times a number

May you tell me which one is the Correct Answer, Thank you.

nexus9112 [7]

No the correct answer is c. If you use the formula
Y2- Y1
X2-X1
You could use this to find the slope. And then the y-intercept if any.

6 0

3 years ago

ABCD is parallelogram if angle D=32, then angle c=?

Nonamiya [84]

<C = 180 - 32
<C = 148 degrees

(<C+<D = 180)

5 0

3 years ago

Can anyone help me solve these linear systems using substitution?

vivado [14]

The system of the linear systems of equation using substitution is;

x = 2, y = 2
x = -20, y = -1

<h3>Linear equation</h3>

3x-y=4

x+2y=6

from (2)

x = 6 - 2y

substitute into (1)

3x-y=4

3(6 - 2y) - y = 4

18 - 6y - y = 4

- 6y - y = 4 - 18

-7y = -14

y = 2

Substitute into

x+2y=6

x + 2(2) = 6

x + 4 = 6

x = 6 - 4

x = 2

2. 2x-y= -39

x+y= -21

From (2)

x = -21 - y

substitute into

2x-y= -39

2(-21 - y) - y = -39

-42 - 2y - y = -39

- 2y - y = -39 + 42

- 3y = 3

y = 3/-3

y = -1

substitute into

x+y= -21

x + (-1) = -21

x - 1 = -21

x = -21 + 1

x = -20

3. 2x+y =11

6x-5y =9

Learn more about linear equation:

brainly.com/question/4074386

#SPJ1

6 0

1 year ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

Gavin is building a model of a kitchen in the model 2/5 of the floor tiles or white one half of the floor tiles are yellow and o

denis-greek [22]

I think it's 10 because 10 is a multiple of 10,2, and 5. 5x2=10 2x5=10 10x1=10. Hoped I helped:D

7 0

3 years ago

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