plzz help me asap i need help picture below
Which line has a slope of -?



line d
line b
line a
line c

Answer:

It cannot be 0

Step-by-step explanation:

it can also be positive number :2/4

it can be odd number too:3/9

it is bigger than numerator bcoz we have to divide it for numerator

So, 0 number cannot be put as denominator in fraction is true statement

Answer:

It is a ray because there are two points with a line passing through them which is extenging on one side but not on the other.

Tanθ =sin /cos

tan θ = 5/2 / y

tan (30°) = 5/2 /y

[tex]y = \frac{5 \sqrt{3} }{2} [/tex]

y=4.33

If y (0) = -1/3, then

-1/3 = 1 / (1 + C e ⁻⁰)

Solve for C :

-1/3 = 1 / (1 + C )

-3 = 1 + C

C = -4

So the particular solution to the DE that satisfies the given initial condition is

[tex]\boxed{y=\dfrac1{1-4e^{-x}}}[/tex]

Answer:

[tex]\angle P[/tex]

Step-by-step explanation:

Given

[tex]\triangle PRQ = \triangle TSU = 90^o[/tex]

[tex]PQ = 20[/tex]     [tex]QR = 16[/tex]    [tex]PR = 12[/tex]

[tex]ST = 30[/tex]       [tex]TU = 34[/tex]    [tex]SU = 16[/tex]

See attachment

Required

Which sine of angle is equivalent to [tex]\frac{4}{5}[/tex]

Considering [tex]\triangle PQR[/tex]

We have:

[tex]\sin(P) = \frac{QR}{PQ}[/tex] --- i.e. opposite/hypotenuse

So, we have:

[tex]\sin(P) = \frac{16}{20}[/tex]

Divide by 4

[tex]\sin(P) = \frac{4}{5}[/tex]

Hence:

[tex]\angle P[/tex] is correct

Answer:

A or <P

Step-by-step explanation:

on edge 2021

Answer:

C

Step-by-step explanation:

The graph has asymptotes at x = 2 and x = -1 corresponding to the denominator of option C.

Answer:

5:12

Step-by-step explanation:

125:300 simplified = 5:12

I hope this helps

Answer:

21

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 60 = x / 7 sqrt(3)

7 sqrt(3)  tan 60 = x

7 sqrt(3) sqrt(3) = x

7*3 = x

21 = x

9514 1404 393

Answer:

  (b)  0 = -78

Step-by-step explanation:

Subtracting 6 times the first equation from the second will give ...

  (4x +15y) -6(2/3x +5/2y) = (12) -6(15)

  0 = -78

Answer:

the answer is b

Step-by-step explanation:

Answer:

(-2,2) and (-3,-4)

Step-by-step explanation:

by graphing the line and parabola, you should get this graph

Answer:

0.26684

Step-by-step explanation:

Given that :

Mean, μ = 62.5

Standard deviation, σ = 1.96

P(Z ≥ 63.72)

The Zscore = (x - μ) / σ

P(Z ≥ (x - μ) / σ)

P(Z ≥ (63.72 - 62.5) / 1. 96) = P(Z ≥ 0.6224)

P(Z ≥ 0.6224) = 1 - P(Z < 0.6224)

1 - P(Z < 0.6224) = 1 - 0.73316 = 0.26684

Answer:

dC=3.5

DC is between 3.475 and 3.525

Step-by-step explanation:

So let dx=1 since the change there is a change in 1 unit.

Find dC/dx by differentiating the expression named C.

dC/dx=0.05x+3

So dC=(0.05x+3) dx

Plug in x=10 and dx=1:

dC=(0.05×10+3)(1)

dC=(0.5+3)

dC=3.5

Let D be the change in cost-the triangle thing.

Since dx=1 we only want the change in unit to be within 1 in difference.

So this means we want it to be from x=9 to x=1] ot from x=10 to x=11.

Let's do from x=9 to x=10 first:

DC=C(10)-C(9)

DC=[0.025×10^2+3×10+4]-[0.025×9^2+3×9+4]

DC=[2.5+30+4]-[0.025×81+27+4]

DC=[36.5]-[2.025+31]

DC=[36.5]-[33.025]

DC=3.475

Now let's do from x=10 to x=11

DC=[0.025×11^2+3×11+4]-[0.025×10^2+3×10+4]

DC=[0.025×121+33+4]-[36.5]

DC=[3.025+37]-[36.5]

DC=[40.025]-[36.5]

DC=3.525

So DC, the change in cost where the change in unit is 1, is between 3.475 and 3.525.

Answer:

The slope is undefined.

Step-by-step explanation:

The line must pass through the points (10,-3) and (10,-8), meaning that it must be vertical. The slope of a line is undefined if the line is vertical.

Answer:

6

Step-by-step explanation:

236/10 = 23 remainder 6, so 6 crayons is the answer

Answer:

x=60

Step-by-step explanation:

There are different ways to find x but this is what I found easiest.

To solve first note that AOD and CFB are vertical angles; this means that they are congruent. AOD consists of two angles with the measurements of 90 and x. CFB consists of two angles with the measurements of 30 and 2x. So, to find x set add the adjacent angles and set them equal to the other pair of angles. The equation would be [tex]90+x=30+2x[/tex]. First, subtract x from both sides; this makes the equation [tex]90=30+x[/tex]. Then, subtract 30 from both sides. This gives the final answer, x=60.

Answer:

A one solution

Step-by-step explanation:

4(x - 5) = 3x + 7

Distribute

4x - 20 = 3x+7

Subtract 3x from each side

4x-3x-20 = 3x+7-3x

x -20 = 7

Add 20 to each side

x -20+20 = 7+20

x = 27

There is one solution

Answer:

Step-by-step explanation:

Let's simplify that before we make the decision, shall we? We'll get rid of the parenthesis by distribution and then combine like terms.

4x - 20 = 3x + 7 and combining like terms and getting everything on one side of the equals sign:

1x - 27 = 0. Since that x has a power of 1 on it (linear), that means we have only 1 solution. If that was an x², we would have 2 solutions; if that was an x³, we would have 3 solutions, etc.

Answer:

D: Start at the origin. Move 3 and one-half units left because the x-coordinate is Negative 3 and one-half. Negative 3 and one-half is between -3 and -4. Move 2 units down because the y-coordinate is -2.

Answer:

20'×15 in 54 inches

Step-by-step explanation:

The Best as a pool should be rectangular in shape and 54inches deep for safety of life's

The volume of the rectangular pool that is 20' x 15' in 54 inches deep is largest.

The volume of the cylinder is the product of the height, pie, and square of the radius.

The volume of the cylinder = [tex]\pi r^{2}[/tex]h

The volume of the cylindrical pool that has a 3.3 m radius and is 1.8 m deep is;

=  [tex]\pi r^{2}[/tex]h

[tex]= 3.14 (3.3)^2 (1.8)\\\\= 61.55 m^3[/tex]

The volume of the rectangular pool that is 20' x 15' in 54 inches deep ;

V = 20 x 15 x 54

V = 16,200 cubic meter.

The volume of the rectangular pool that is 20' x 15' in 54 inches deep is largest.

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Answer:

14cm²

Step-by-step explanation:

3x2=6,

3x2=6,

2x1=2,

6+6+2=14 cm^2

Answer:

jhshejwjabsgsgshshsnsjs

Answer:

Step-by-step explanation:

Put the equation into center-radius form.

x² + y² + 8x - 12y + 24 = 0

x² + y² + 8x - 12y = -24

(x²+8x) + (y²-12y) = -24

(x²+8x+4²) + (y²-12y+6²) = 4²+6²-24

(x+4)² + (y-6)² = 28

Center: (-4,6)

radius: √28

Answer:

[tex]a = 4, -4[/tex]

Step-by-step explanation:

Step 1:  Plug in -4 for c

[tex]-a^{2} + c^{2}[/tex]

[tex]-a^{2} + (-4)^{2}[/tex]

[tex]-a^{2} + 16[/tex]

Step 2:  Solve for a

[tex]-a^{2}+16-16=0-16[/tex]

[tex]-a^{2}/-1 = -16/-1[/tex]

[tex]a^{2} = 16[/tex]

[tex]\sqrt{a^{2}} = \sqrt{16}[/tex]

[tex]a = 4, -4[/tex]

Answer:  [tex]a = 4, -4[/tex]

qualitative

Step-by-step explanation:

b cos the question is in quality format

Answer:

cutee!

SUP???

Hiii friend :]

cuteee~!

prettyyy

Answer:

y = 3

Step-by-step explanation:

6sin(30) = 3

Answer:

5 bags of cement are required.

Step-by-step explanation:

Since to make concrete, the ratio of cement to sand is 1: 3, if cement and sand are sold in bags of equal mass, to determine how many bags of cement are required to make concrete using 15 bags of sand the following calculation must be done:

Cement = 1

Sand = 3

3 = 15

1 = X

15/3 = X

5 = X

Therefore, 5 bags of cement are required.

Answer:

[tex]{ \bf{f(x) = \tan x + 9 \sin x }}[/tex]

For gradient, differentiate f(x):

[tex]{ \tt{ \frac{dy}{dx} = { \sec }^{2}x + 9 \cos x }}[/tex]

Substitute for x as π:

[tex]{ \tt{gradient = { \sec }^{2} \pi + 9 \cos(\pi ) }} \\ { \tt{gradient = - 8 }}[/tex]

Gradient of tangent = -8

[tex]{ \bf{y =mx + b }} \\ { \tt{0 = ( - 8\pi) + b}} \\ { \tt{b = 8\pi}} \\ y - intercept = 8\pi[/tex]

Equation of tangent:

[tex]{ \boxed{ \bf{y = - 8x + 8\pi}}}[/tex]

Answer:

Step-by-step explanation:

the arrows from the picture tells us that TV is parallel to RS

since TS is a transversal that cuts the 2 parallel lines TV and RS than ∠S =x

(alternate interior angles)

sum of angles in a Δ is 180° so x+x+2x = 180°, 4x =180°, x= 45°

2x = 45*2 = 90°

Answer:

Given E(t)=1100(100-t)+580(100-t)^2

Put t = 99.5, we get

E(99.5)=1100(100-99.5)+580(100-99.5)^2

E(99.5)=1100(0.5)+580(0.5)^2

E(99.5)=1100(0.5)+580(0.25)

E(99.5)=550+145

E(99.5)=695m

Step-by-step explanation:

It can be concluded that -

E(99.5) = 695

E(100) = 0

In mathematics, an expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Mathematical symbols can designate numbers (constants), variables, operations, functions, brackets, punctuation, and grouping to help determine order of operations and other aspects of logical syntax.

Given is the function as follows -

E(t) = 1100(100 - t) + 580(100 - t)²

The given function is -

E(t) = 1100(100 - t) + 580(100 - t)²

At → E(99.5)

E(99.5) = 1100(100 - t) + 580(100 - t)²

E(99.5) = 1100(100 - 99.5) + 580(100 - 99.5)²

E(99.5) = 1100(0.5) + 580(0.5)²

E(99.5) = 550 + 145

E(99.5) = 695

At → E(100)

E(100) = 1100(100 - t) + 580(100 - t)²

E(100) = 1100(100 - 100) + 580(100 - 100)²

E(100) = 0

Therefore, it can be concluded that -

E(99.5) = 695

E(100) = 0

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Answer: $40 / $80

Step-by-step explanation: 40$ if it's $8 for BOTH per hour, or if it's $8 for ONE per hour it's $80

Answer:

[tex]\displaystyle 51[/tex]

General Formulas and Concepts:

Algebra I

Algebra II

Calculus

Limits

Limit Rule [Variable Direct Substitution]:                                                             [tex]\displaystyle \lim_{x \to c} x = c[/tex]

Limit Property [Addition/Subtraction]:                                                                   [tex]\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)[/tex]

Limit Property [Multiplied Constant]:                                                                     [tex]\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle f(x) = \left \{ {{3\sqrt{2x - 1}, \ x \leq 2} \atop {6(2x - 1)^2 - 3, \ x > 2}} \right.[/tex]

Step 2: Solve

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Book: College Calculus 10e