All categories

3 years ago

Someone help plss show work

Mathematics

2 answers:

yanalaym [24]3 years ago

8 0

Answer:

Slope: 3/5

Y int: (0,-5)

Step-by-step explanation:

Mazyrski [523]3 years ago

4 0

Answer:

Slope: $\frac{3}{5}$

Y-Intercept: $(0,-5)$

Step-by-step explanation:

1. To find the slope, we need to have the equation in slope-intercept form, which is $y=mx+b$ . In this case, our equation is already in this form. $m$ is what your slope is. In this case, our

2. To find the y-intercept we look at value $b$ that is in $y=mx+b$ . In this case, $b$ is $-5$ . So, our ordered pair is $(0,-5)$ .

Read more about reflection at:

brainly.com/question/4289712

4 0

2 years ago

If you rotated the shape around the line, which three-dimensional figure would be formed?

Fiesta28 [93]

Answer:

c cylinder

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Help pleaseeeee ???????

Mkey [24]

Answer: y+6x = -34

Step-by-step explanation:

y-2= -6(x+6) (Ax+By=C is the goal)

y-2= -6x-36 (-6) x & (-6) 6

y=-6x-34 (Add 2 to both sides)

y+6x = -34 [Add 6x to both sides]

5 0

3 years ago

E3: Amab sells an article to Vivek at a profit of 15% and Vivek sells it to

Bogdan [553]

Answer:

the cost price is 27,715

Step-by-step explanation:

Let the cost price be x

As Amab sells to Vivek at 15% profit

So the selling price be

= 115 ÷ 100 × x = 1.15x

And Vivek sells to sagar at a profit of 10%

= 1.15x × 110 ÷ 100

= 1.265x

Now sagar pays it for 35,060

So , the cost price be

1.265x = 35,060

x = 27,715

Hence, the cost price is 27,715

3 0

3 years ago

Determine the equation of the line that is perpendicular to the lines r(t) equalsleft angle6t,1 plus3 t,4tright angle and R(

Lynna [10]

$\vec r(t)=\langle6t,1+3t,4t\rangle$

$\vec R(s)=\langle2+s,-8+3s,-12+4s\rangle$

Take the derivatives of each to get the tangent vectors:

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle6,3,4\rangle$

$\dfrac{\mathrm d\vec R(s)}{\mathrm ds}=\langle1,3,4\rangle$

Take the cross product of the tangent vectors to get a vector that is normal to both lines:

$\langle6,3,4\rangle\times\langle1,3,4\rangle=\langle0,-20,15\rangle$

The two given lines intersect when $\vec r(t)=\vec R(s)$ :

$\langle6t,1+3t,4t\rangle=\langle2+s,-8+3s,-12+4s\rangle\implies t=1,s=4$

that is, at the point (6, 4, 4).

The line perpendicular to both of the given lines through the origin is obtained by scaling the normal vector found earlier by $\tau\in\Bbb R$ ; translate this line by adding the vector $\langle6,4,4\rangle$ to get the line we want,

$\vec\rho(\tau)=\langle6,4,4\rangle+\langle0,-20,15\rangle\tau$

$\boxed{\vec\rho(\tau)=\langle6,4-20\tau,4+15\tau\rangle}$

6 0

3 years ago

Someone help plss show work​

Someone help plss show work