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2 years ago

7

The difference in the x- coordinates of two points is 3, and the difference in the y- coordinates of the two points is 6. What i

s the slope of the line that passes through the points?

2 answers:

USPshnik [31]2 years ago

7 0

Answer:

2

Step-by-step explanation:

Slope or gradient of a line is given by dividing change in y axis by the change in x axis. Slope is represented by m hence $m=\frac {\triangle y}{\triangle x}$ where $\triangle y$ is the change in y axis while $\triangle x$ is the change in x co-ordinates. Substituting $\triangle y$ with 6 units while 3 units substitute $\triangle x$ then the slope will be $m=\frac {6}{3}=2$

Therefore, the gradient is 2

Nikolay [14]2 years ago

3 0

Answer:

A. 2

Step-by-step explanation:

I got it right on Edgenuity 2020!

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You might be interested in

The population of Smalltown in the year 1890 was 6,250. Since then, it has increased at a rate of 3.75% each year. • What was th

Vikki [24]

Answer:

year 1915 : 15,689

year 1940 : 39,381

t = 137.9 years

Step-by-step explanation:

Hi, to answer this question we have to apply an exponential growth function:

A = P (1 + r) t

Where:

p = original population

r = growing rate (decimal form)

t= years

A = population after t years

Replacing with the values given:

A = 6,250 (1 + 3.75/100)^t

A = 6,250 (1 + 0.0375)^t

A = 6,250 (1.0375)^t

So, by the year 1915 :

1915-1890 = 25 years passed (t)

A = 6,250 (1.0375)^25

A = 15,689

by the year 1940 :

1940-1890 = 50 years passed (t)

A = 6,250 (1.0375)^50

A = 39,381

When will the population reach 1,000,000?. We have to subtitute A=1000000 and solve for t.

1,000,000= 6,250 (1.0375)^t

1,000,000/ 6,250 =(1.0375)^t

160 = 1.0375^t

log 160 = log 1.0375^t

log 160 = (t ) log 1.0375

log160 / log 1.0375= t

t = 137.9 years

8 0

2 years ago

Which graph represents the function y = 2x - 4

WARRIOR [948]

Answer:

I will change my awnser when you tell me but what are the awnser choices

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students re

Vladimir79 [104]

Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ = 0.95

The 95% of confidence intervals

$x^{-} ± t_{\alpha } \frac{S}{\sqrt{n} }$

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} },x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })$

$(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )$

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)

4 0

2 years ago

Find the dimensions of the box with volume 1728 cm3 that has minimal surface area. (Let x, y, and z be the dimensions of the box

Gala2k [10]

Answer:

The dimensions of the box are 12 cm , 12 cm , 12 cm

Step-by-step explanation:

Let x , y and z be the dimensions of box

Volume of box =xyz=1728

$z=\frac{1728}{xy}$

Surface area of box = $2xy+2yz+2xz=2xy+2y(\frac{1728}{xy})+2x(\frac{1728}{xy})$

Let $f(x,y)=2xy+2(\frac{1728}{x})+2(\frac{1728}{y})$

To get minimal surface area

$\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial x}=0$

$2y-2(\frac{1728}{x^2})=0$

$y=\frac{1728}{x^2}$ ----1

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial y}=0$

$2x-2(\frac{1728}{y^2})=0\\x=\frac{1728}{y^2} \\y^2=\frac{1728}{x}$

Using 1

$(\frac{1728}{x^2} )^2=\frac{1728}{x}$

x=0 and $x^3=1728$

Side can never be 0

So, $x^3=1728$

x=12

$y=\frac{1728}{x^2} \\y=\frac{1728}{12^2}$

y=12

$z=\frac{1728}{xy}\\z=\frac{1728}{(12)(12)}$

z=12

The dimensions of the box are 12 cm , 12 cm , 12 cm

5 0

2 years ago

A cake manufacturer boxes Swiss chocolate and German chocolate cakes at one site. A box of Swiss chocolate cake contains 0.55 lb

Kobotan [32]

Answer:

2,500 German chocolate cake boxes.

1,500 Swiss chocolate cake boxes.

Step-by-step explanation:

Let 'S' be the number of Swiss chocolate cakes boxed and 'G' the number of German cholocate cakes boxed. If all of the available ingredients are used:

$0.55*S +0.59*G = 2,300\\0.45*S+0.41*G = 1,700$

Solving the linear system above:

$(0.55 *S +0.59*G)- \frac{0.55}{0.45} (0.45*S+0.41*G)= 2,300-\frac{0.55}{0.45}*1,700*\\0.088888*G = 222.222\\G=2,500\\S =\frac{2,300-2,500*0.59}{0.55}\\S=1500$

2,500 German chocolate cake boxes and 1,500 Swiss chocolate cake boxes can be made each day.

6 0

2 years ago