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3 years ago

Need help with these. 11th grade algebra 2 class. thanks

Mathematics

1 answer:

OlgaM077 [116]3 years ago

3 0

15) x=31/10
16)x=2
17)x=16/3
18)x=-2
19)2.5 hours
20) 32 hours

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Let f be a function of two variables that has continuous partial derivatives and consider the points

Sergeu [11.5K]

Answer:

The directional derivative of f at A in the direction of $\vec{u}$ AD is 7.

Step-by-step explanation:

Step 1:

Directional of a function f in direction of the unit vector $\vec{u}=(a,b)$ is denoted by $D\vec{u}f(x,y)$ ,

$D\vec{u}f(x,y)=f_{x}\left ( x ,y\right ).a+f_{y}(x,y).b$ .

Now the given points are

$A(8,9),B(10,9),C(8,10) and D(11,13)$ ,

Step 2:

The vectors are given as

AB = (10-8, 9-9),the direction is

$\vec{u}_{AB} = \frac{AB}{\left \| AB \right \|}=(1,0)$

AC=(8-8,10-9), the direction is

$\vec{u}_{AC} = \frac{AC}{\left \| AC \right \|}=(0,1)$

AC=(11-8,13-9), the direction is

$\vec{u}_{AD} = \frac{AD}{\left \| AD \right \|}=\left (\frac{3}{5},\frac{4}{5} \right )$

Step 3:

The given directional derivative of f at A $\vec{u}_{AB}$ is 9,

$D\vec{u}_{AB}f=f_{x} \cdot 1 + f_{y}\cdot 0\\f_{x} =9$

The given directional derivative of f at A $\vec{u}_{AC}$ is 2,

$D\vec{u}_{AB}f=f_{x} \cdot 0 + f_{y}\cdot 1\\f_{y} =2$

The given directional derivative of f at A $\vec{u}_{AD}$ is

$D\vec{u}_{AD}f=f_{x} \cdot \frac{3}{5} + f_{y}\cdot \frac{4}{5}$

$D\vec{u}_{AD}f=9 \cdot \frac{3}{5} + 2\cdot \frac{4}{5}$

$D\vec{u}_{AD}f= \frac{27+8}{5} =7$

The directional derivative of f at A in the direction of $\vec{u}_{AD}$ is 7.

3 0

3 years ago

When solved correctly, the equation 4x−8−−−−−√=x−5 has solutions of x=3 and x=11.

grigory [225]

Answer:

x = 3 is extraneous

Step-by-step explanation:

Given

$\sqrt{4x-8}$ = x - 5 ( square both sides to clear the radical )

4x - 8 = (x - 5)² ← expand using FOIL

4x - 8 = x² - 10x + 25 ( subtract 4x - 8 from both sides )

0 = x² - 14x + 33 ← in standard form

0 = (x - 11)(x - 3) ← in factored form

Equate each factor to zero and solve for x

x - 11 = 0 ⇒ x = 11

x - 3 = 0 ⇒ x = 3

As a check

Substitute these values into the equation and if both sides are equal then they are solutions

x = 11

left side = $\sqrt{44-8}$ = $\sqrt{36}$ = 6

right side = 11 - 5 = 6

Thus x = 11 is a solution

x = 3

left side = $\sqrt{12-8}$ = $\sqrt{4}$ = 2

right side = 3 - 5 = - 2 ≠ 2

Thus x = 3 is an extraneous solution

7 0

3 years ago

PLS HELPP Help Solve for x <br>X =

Basile [38]

Answer:

x = 9

Step-by-step explanation:

The big triangle is similar to smaller one so we can use the side lengths' proportion to find x:

7/4 = (x-3)+8/8

7/4 = (x+5)/8 cross multiply expressions

56 = 4x + 20 subtract 20 from both sides

36 = 4x divide both sides by 4

9 = x

4 0

3 years ago

over the weekend Mario rode 43 miles. On Sunday he rode 3 miles more than 2/3 his distance on Saturday. How many miles did he ri

Ierofanga [76]

X+ (2x/3 +3) = 43
$3x + 2x + 9 = 43$
$5x + 9 = 43$
$5x = 34$
$x = \frac{34}{5}$

8 0

4 years ago

posting this again, pls help with c and d and explain thoroughly this is due in an hour i’ll give u brainliest, no links,

anygoal [31]

Answer:

C) a = 6 D) m = 5

Step-by-step explanation:

C) 3a + 11 = 7a - 13

4a = 24

a = 6

D) 2m + m - 8 = 7

3m = 15

m = 5

5 0

3 years ago

Read 2 more answers

Need help with these. 11th grade algebra 2 class. thanks ​

Need help with these. 11th grade algebra 2 class. thanks