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3 years ago

What is a linear function on a graph

2 answers:

7 0

In mathematics, the term linear function refers to two distinct but related notions: In calculus and related areas, a linear function is a function whose graph is a straight line, that is a polynomial function of degree one or zero

JulijaS [17]3 years ago

6 0

Answer:

Step-by-step explanation:

A straight line that has one independent variable and one dependant variable.

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WILL MARK BRAINLIEST! Can someone please help! I don't understand some of these questions :(

PolarNik [594]

Answer:

Step-by-step explanation:

The interior and exterior angle of a polygon is supplementary

let interior be I

let exterior be E

I + E = 180

Since the interior angle is 8 times that of an exterior angle,

8E + E = 180 [replacing I with 8E]

9E = 180

E = 20

The exterior angle is 20 degrees

I + E = 180

I + 20 = 180

I = 160

The interior angle is 160 degrees.

The equation to find the interior angle of a polygon with 'n' number of sides is:

I = ( (n − 2) × 180 ) ⁄ n

We know the interior angle, so plug it in and solve for n:

160 = ( (n − 2) × 180 ) ⁄ n

160n = (n − 2) × 180

160n = 180n − 360

-20n = -360

n = 18

3 0

3 years ago

Nate and his sister Natalie are walking around the track at school and they’re walking at a steady rate. Nate walks 5 feet in 2

lutik1710 [3]

Nate Natalie
5_2 2_2
10_4 4_4
15_6 6_6

Natalie walked 6 feet

7 0

3 years ago

Read 2 more answers

Which of these points is farthest from the origin?

yarga [219]

Answer:

The answer to the picture is 19 square units if I remember correctly.

Step-by-step explanation:

4 0

3 years ago

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

A cylinder with a 2 centimeter diameter is drilled through the triangular prism shown on the left find the volume of the prism

saw5 [17]

Answer:

The volume of a prism on its own = 248.5cm^3

The volume of a cylinder = 87.92cm^3

Step-by-step explanation:

The volume does not change for the prism as the hole was drilled.

We can show the drilled space with v(l) and (p) for prism V

v(l) 248.5p - 87.92p + 87.92p

= 248.5p

3 0

3 years ago