Ask question

Ask question

All categories

2 years ago

14

Helllllllllllllllllpppppppppppppp

1 answer:

pentagon [3]2 years ago

4 0

Answer:

its the second one ........

You might be interested in

Write the equation in slope-intercept form of the line that passes through

mixer [17]

Answer:

We conclude that the equation in slope-intercept form of the line that passes through (12,9) and is perpendicular to the graph of y = -3/4x + 1 will be:

$y=\frac{4}{3}x-7$

Step-by-step explanation:

We know the slope-intercept form of the line equation

y = mx+b

where

m is the slope

b is the y-intercept

Given the line

y = -3/4x + 1

comparing with the slope-intercept form of the line equation

The slope = m = -3/4

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = -3/4

Thus, the slope of the new perpendicular line = – 1/(-3/4) = 4/3

Using the point-slope form

$y-y_1=m\left(x-x_1\right)$

where m is the slope of the line and (x₁, y₁) is the point

substituting the values of the slope = 4/3 and the point (12, 9)

$y-y_1=m\left(x-x_1\right)$

$y-9=\frac{4}{3}\left(x-12\right)$

Add 9 to both sides

$y-9+9=\frac{4}{3}\left(x-12\right)+9$

$y=\frac{4}{3}x-16+9$

$y=\frac{4}{3}x-7$

Therefore, we conclude that the equation in slope-intercept form of the line that passes through (12,9) and is perpendicular to the graph of y = -3/4x + 1 will be:

$y=\frac{4}{3}x-7$

7 0

2 years ago

Ihorangi travels 51 km to work and 51 km from work each day.

viktelen [127]

Answer:

Ihorangi uses 7.65 litres of petrol to and from work

Step-by-step explanation:

If the car Ihorangi drives consumes 7.5 L/100 km

Then for 1 km journey, the car will consume 7.5/100 L

= 0.075 L

If Ihorangi travels 51 km to work and 51 km from work

Therefore Ihorangi travel (51 + 51) km daily to and from work

= 102 km daily

The fuel he consumes daily is 102*0.075 L

= 7.65 L

7 0

3 years ago

In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t an

Alenkasestr [34]

Answer: 693 years

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

$A(t) =A_0e^{kt}$

k = rate constant

t = time taken for decomposition = 1

$A_0$ = Initial amount of the reactant

$A_t$ = amount of the reactant left = $A_0-\frac{0.1}{100}\times A_0=0.999A_0$

$0.999A_0=A_0e^{k\times 1}$

$0.999=e^k$

$k=-0.001year^{-1}$

for half life : $t=t_\frac{1}{2}$

$A_t=\frac{1}{2}A_o$

Putting in the values , we get

$\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}$

$t_\frac{1}{2}=693 years$

Thus half life of this isotope, to the nearest year is 693.

8 0

3 years ago

When completely factored, x^2 + x - 6 is equivalent to which of the following? A. (x + 1)(x – 6) B. (x + 3)(x – 2) C. (x + 6)(x

Mariana [72]

X^2+x-6

x^2-2x+3x-6

x(x-2)+3(x-2)

(x+3)(x-2)

5 0

3 years ago

Which transformations will produce similar, but not congruent, figures?

n200080 [17]

A, because anything with a dilation will not be congruent as the original PQR :))

7 0

2 years ago

Read 2 more answers