HURRYY Find the rate of change for the ramp represented in the graph

HELP PLEASE!Write the equation of the line parallel to y = –x + 5 passing through the point (9, –1).

Westkost [7]

Y=-x+8
gradient of the parallel line is the same as the gradient of the line given, which is -1
to find the y-intercept, sub the values (9,-1) into the equation we are finding which is y=-x+c

5 0

2 years ago

Two bugs run along the rectangular starting from the same corner. The bug who runs east is two inches per second faster than the

german

Answer:

6 and 8

Step-by-step explanation:

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cfrak%7B%20%5Corange%7B%20Question%7D%7D" id="TexFormula1" title=" \large\frak{ \

SVEN [57.7K]

Answer:

$\boxed{\sf distance \: between \: the \: objects \: \: is \: 2.5 \: cm}$

Step-by-step explanation:

According to universal law of gravitation every object in the universe attracts every other particles that surrounds it with a force which is inversely proportional to the square of their distance of separation & directly proportional to

the product of their masses, given by standard formula.

$A = G \cdot \frac{m_1.m_2}{ {d}^{2} }$

where A,d & G are force of attraction, distance of separation & proportionality constant respectively.

Given:

A1 = 2 units

d1= 5 cm

A2 = 8 units

To find:

Distance of separation when the force of attraction is 8 units d2 = ?

Solution:

Substituting the given values in above at each point,

$A_1 = G \cdot \frac{m_1.m_2}{ {d_1}^{2} } \\ 2 = G \cdot \frac{m_1.m_2}{ {5}^{2} } \\ \sf similarly \: at \: second \: point \: of \: attraction \\ A_2 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ 8 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ \sf \: \: dividing \: both \: of \: the \: equation \\ \frac{2}{8} = \frac{G \cdot \frac{m_1.m_2}{ {5}^{2} }}{G \cdot \frac{m_1.m_2}{ {d_2}^{2} } } \\ \sf \: G \: m_1 and \: m_2 \: are \: \: constant \: hence \\ \sf they \: can \: be \: cancelled \: out \\ \frac{2}{8} = \frac{ \frac{1}{ {5}^{2} } }{ \frac{1}{ {d_2}^{2} } } \\ \sf rearranging \: above \: equation \\ \frac{1}{4} = \frac{{d_2}^{2} }{ {5}^{2} } \\ {d_2}^{2} = \frac{ {5}^{2} }{4 } \\ {d_2}^{2} = \frac{25}{4} \\ {d_2}^{2} = 6.25 \\ \sqrt{{d_2}^{2} } = \sqrt{6.25} \\ \boxed{ \sf{d_2} = 2.5 \: cm}$