All categories

3 years ago

I need help please, very confused on what to do!

Mathematics

1 answer:

Firdavs [7]3 years ago

3 0

Step-by-step explanation:

X intercept of -5 means the line cuts the x axis at -5 so, mark with a small dot/x at - 5 on the x axis which is the horizontal line

y intercept of 3 means the line cuts/passes thru the y axis at 3. so, mark it with a dot/cross at 3 on the vertical line

now, join the 2 markings together with a straight line

You might be interested in

X(x) + 12x + 21 = 0 factor

Yakvenalex [24]

It is not factorable. Factors of 21 are. 21 and 1
7 and 3. None of those add up to 12

3 0

3 years ago

a fair sold a total of 3,300 tickets on Friday and Saturday. It sold 100 more on Friday than on Saturday. How many tickets did t

xz_007 [3.2K]

3200 is the answer thank you

6 0

3 years ago

Find the number of pens

Semenov [28]

There are 44 pens in all

5 0

3 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago

A watercolor painting is 24 inches long by 11 inches wide. Ramon makes a border around the watercolor painting by making a mat t

sergeinik [125]

Given: It is given that the length of the painting is 24 inches and the width is 11 inches.

To find: Area of the mat

Solution:

The watercolor painting is 24 inches long by 11 inches wide.

So, the area of the painting is:

$\text{Area of painting}=l\times b$

$\text{Area of painting}=24\times 11$

$\text{Area of painting}=264 \text{ in}^2$

The length of painting with mat is = 24 in + 3 in + 3 in = 30 in

The width of painting with mat = 11 in + 3 in + 3 in = 17 in

$\text{Area of painting with mat}=30\times 17$

$\text{Area of painting with mat}=510 \text{ in}^2$

Now to calculate the area of mat subtracts the area of painting from the area of the mat.

$\text{Area of mat}=\text{Area of painting with mat}-\text{Area of painting}\\$

$\text{Area of mat}=510-264$

$\text{Area of mat}=246 \text{ in}^2$

Hence, the area of the mat is 246 in².

6 0

2 years ago