All categories

2 years ago

Helpppppppppppppppppppppppp WILL MARK BRAINLIST

Mathematics

1 answer:

Brilliant_brown [7]2 years ago

7 0

Sub in -7 as x
-(-7)^2 + 8(-7)
= -49 -56
= -105

You might be interested in

Using appropriate properties find: 2/3 × 3/4 + 3/7 × 3/4

Viktor [21]

Answer:

$\frac{23}{28}$

Step-by-step explanation:

$\frac{2}{3} \times \frac{3}{4} + \frac{3}{7} \times \frac{3}{4}$

$= > \frac{3}{4} ( \frac{2}{3} + \frac{3}{7} )$

$= > \frac{3}{4} \times \frac{23}{21}$

Reducing 3 from numerator and denominator,

$= > \frac{23}{4 \times 7} = \frac{23}{28}$

3 0

2 years ago

Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

fredd [130]

Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

Second tangent line:

$y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7$

Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

6 0

2 years ago

Read 2 more answers

Jessica wants to buy a purse that costs $25.00. She has already saved

jekas [21]

Answer:

Step-by-step explanation:

19-25

4 0

2 years ago

Read 2 more answers

Write an equation of a parabola with a vertex at the origin and the given focus 10. Focus at (-1,0)

Nostrana [21]

Step-by-step explanation:

focus(p) is on the x axis= -1

vertex=(0,0)

parabola equation----- (y-h)^2 =4p(x-k)

(h, k)=(0,0)

(y-0)^2 =4(-1)(x-0)

y^2 = -4x

that is the answer

3 0

2 years ago

Can someone please please help me on this algebra problem I’m stuck on answering the question?

Rudik [331]

Explanation:

Midline: it is a horizontal line where half of the function above it and half of the function is below it.

Amplitude: It means how far the function varies from the midline to the above and below.

Period: the distance between two consecutive maximum points (minimum points).

3) From the graph,

$\begin{gathered} \text{Midline}=\frac{Maximum\text{ point+ minimum point}}{2} \\ =\frac{3+(-1)}{2} \\ =\frac{3-1}{2} \\ =\frac{2}{2} \\ =1 \end{gathered}$

Answer: Midline=1

1) Amplitude

$\text{Amplitude =}2$

2) Periods

$\begin{gathered} \text{Period}=6 \\ \Rightarrow b=\frac{2\pi}{6} \end{gathered}$

4) The trigonometric equation is,

$\begin{gathered} y=a\cos (b(x-c))+d \\ a=\text{ amplitude} \\ b=\text{period} \\ c=\text{ phase shift} \\ d=\text{ vertical shift} \\ y=2\cos (\frac{2\pi}{6}x)+1 \end{gathered}$

Answer:

$y=2\cos (\frac{2\pi}{6}x)+1$

4 0

1 year ago