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3 years ago

Can someone help me on 11 and 14

Mathematics

1 answer:

sveticcg [70]3 years ago

5 0

For 11, set them equal to each other, since they both equal y.

y=2x+5; y=3x-1
2x+5=3x-1
x=6

and then substitute that x-value to find y.

y=3(6)-1
y=18-1
y=17

for 14 you can substitute the bottom y-value (x+7) into the top equation.

2x=y-10; x+7=y
2x=(x+7)-10
x=-3

and then substitute that x-value to find y.

-3+7=y
y=4

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Simplify the expression √6 (√33=7)

marin [14]

14.07 is the answer to the question above

7 0

2 years ago

On a coordinate plane, a line has points (negative 2, negative 4) and (4, 2). Point P is at (0, 4). Which points lie on the line

NikAS [45]

Answer:

the correct options are:

(–1, 3), (–2, 2) and (–5, –1)

Step-by-step explanation:

Given that a line passes through two points

A(-2, -4) and B(4, 2)

Another point P(0, 4)

To find:

Which points lie on the line that passes through P and is parallel to line AB ?

Solution:

First of all, let us the find the equation of the line which is parallel to AB and passes through point P.

Parallel lines have the same slope.

Slope of a line is given as:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{2-(-4)}{4-(-2)} = 1$

Now, using slope intercept form ( $y = mx+c$ ) of a line, we can write the equation of line parallel to AB:

$y =(1)x+c \Rightarrow y = x+c$

Now, putting the point P(0,4) to find c:

$4 = 0 +c \Rightarrow c = 4$

So, the equation is $\bold{y=x+4}$

So, the coordinates given in the options which have value of y coordinate equal to 4 greater than x coordinate will be true.

So, the correct options are:

(–1, 3), (–2, 2) and (–5, –1)

8 0

2 years ago

Read 2 more answers

Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu

kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

$f(x) = {x}^{5} - {x}^{3} + 6$

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

$f'(x) = 5 {x}^{4} - 3 {x}^{2}$

At turning point f'(x)=0.

$5 {x}^{4} - 3 {x}^{2} = 0$

This implies that,

${x}^{2} (5 {x}^{2} - 3) = 0$

${x}^{2} = 0 \: or \: 5 {x}^{2} - 3 = 0$

$x = - \frac{ \sqrt{15} }{5} \: or \: x = 0 \: \: or \: x =\frac{ \sqrt{15} }{5}$

We plug this values into the original function to obtain the y-values of the turning points

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} +750)) \:and \: (0, - 6) \: and\: ( \frac{ \sqrt{15} }{5} , \frac{1}{125} ( - 6 \sqrt{15} +750))$