All categories

1 year ago

The equation f(x)=x−5 translated 4 units to the left is g(x)=

Mathematics

1 answer:

elixir [45]1 year ago

3 0

The equation of the function g(x) is g(x) = x - 1

<h3>How to determine the equation of the function g(x)?</h3>

The given parameters are:

f(x) = x - 5

The above function f(x) is translated left by 4 units

So, we have

g(x) = f(x + 4)

This gives

f(x + 4) = x + 4- 5

Evaluate the sum

f(x + 4) = x - 1

So, we have:

g(x) = x - 1

Hence, the equation of the function g(x) is g(x) = x - 1

Read more about function transformation at:

brainly.com/question/10904859

#SPJ1

You might be interested in

(01.01) Kisha and Rae are asked to solve −5x − 14 = 2x + 7. Identify where one of them made an error.

lora16 [44]

It's A.
Rae made the error when she added 7.
the equation should be:
<span>-14 - 7 = 7x + 7 - 7 </span>

3 0

3 years ago

Read 2 more answers

Solve for n. <br> n<br> ------------ = -2<br> n-12.168

Andrej [43]

Answer:

do we have to put an whole number there

6 0

2 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST !! 100 POINTS

muminat

Answer:

6280 mm squared

Step-by-step explanation:

To find the area Of a cylender we multiply pi times r squared times the height, and we can put pi as 3.14

the radius is 10 mm, and the height is 20 mm.

pug it in.

3.14 times 10 squared times 20.

3.14 time 100 time 20

314 times 20

6280 mm squared

8 0

2 years ago

Read 2 more answers

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

Find the simple interest on $2,219 principal, deposited for 6 years at a rate of 1.91%

Aleks04 [339]

Answer:

$13,568.30

Step-by-step explanation:

I = Prt

I = 2,219(1.0191)(6) = 13,568.30

6 0

2 years ago