What equation correctly applies the quadratic formula?

Answer:

6

Step-by-step explanation:

i took the test

The height of the prism is 49.2 feet

The given parameters are:

The area of the trapezoid is:

Area = 0.5 * (Sum of opposite sides) * Height

So, we have:

1280 = 0.5 * 52 * Height

This gives

1280 = 26 * Height

Divide both sides by 26

Height = 49.2

Hence, the height of the prism is 49.2 feet

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Answer:

Step-by-step explanation:

Hello,

First of all, let s take x real number different from 5, as we cannot divide by 0

[tex]\dfrac{x+5}{x-5}\geq 0 \\ \\<=> (x+5\geq 0 \ and \ x-5> 0) \ or \ (x+5\leq 0 \ and \ x-5< 0) \\ \\<=> (x\geq -5 \ and \ x> 5) \ or \ (x\leq -5 \ and \ x< 5)\\ \\<=> x> 5 \ or \ x\leq -5[/tex]

So the solution is

[tex]\Large \boxed{\sf \bf ]-\infty;-5]\cup]5;+\infty[}[/tex]

Thanks

Answer:

(−∞,−5]∪(5,∞)

Answer:

Step-by-step explanation:

Opposite sides of a parallelogram are congruent, so:

[tex]8x+3=19 \longrightarrow x=2[/tex]

[tex]27=-y-5 \longrightarrow y=-32[/tex]

Answer:

65%

Step-by-step explanation:

You start by dividing 641.29 by 986.60 (641.29/986.60). This gets you .65, which as a percentage is 65%

Answer:

Sensor Calibration

Convert between coordinate systems of every sensor, image, vehicle and global coordinate axis system. So data from a sensor can be mapped globally.

Calibration of a camera helps to convert a pixel in an image to

Computer Vision

Step-by-step explanation:

Answer:

x=3

Step-by-step explanation:

In(x-1) = In(6) - In(x)

In(x-1) + In(x) = In(6)

In((x-1)x) = In(6)

In(x^2-x) = In(6)

x^2-x=6

x^2-x-6=0

x^2+2x-3x-6=0

x(x+2)-3(x+2)=0

(x-3)(x+2)=0

x=3 because x=-2 need to be cancel out

Answer:

z=25.45

because {45.28=19.83+25.45}

{45.28=45.28}

Answer:

B. 75.5 square inches

Step-by-step explanation:

Area (pentagon) = Area (rectangle) - Area (triangle)

= length x width - 1/2 x breadth x height

= 10 x 8 - 1/2 x 3 x 3

= 80 - 4.5

= 75.5 square inches

Solution :-

B. 75.5 square inches

Answer:

B) 75.5 in²

Step-by-step explanation:

Area of rectangle = width × length

⇒ area of rectangle = 8 × 10 = 80 in²

Area of triangle = 1/2 × base × height

⇒ area of triangle = 1/2 × 3 × 3 = 4.5 in²

Area of pentagon = area of rectangle - area of triangle

                              = 80 - 4.5

                             = 75.5 in²

Answer:

Let's solve your equation step-by-step.

p=(0.5)(2.718282)−5

Step 1: Simplify both sides of the equation.

p=(0.5)(2.718282)−5

p=1.359141+−5

p=(1.359141+−5)(Combine Like Terms)

p=−3.640859

p=−3.640859

Answer:

p=−3.640859

Step-by-step explanation:

The volume of the prism that its faces is made from 6 identical squares is 48 cm³.

A square prism is a 3-dimensional cuboid where the base and top are equal squares.

where

a = side length

h = height

Therefore,

a = 2 cm

h = 12 cm

Therefore,

v = a²h

v = 2² × 12

v = 4 × 12

v = 48 cm³

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Answer:

288cm (cubed)

Step-by-step explanation:

4*2=8*12=96

6*2=12*12=144

2*2=4*12=48

144+96+48=288

Answer:

See below

Step-by-step explanation:

Definitions are:

Answer: B 0.4

Step-by-step explanation:

50-20=30

0.4 is 40%

40% of 50 is 20

Answer:

3 hours

Step-by-step explanation:

as jen took 3 hours maryana will take equal amount of time as jen

Well, we already know that it took Jen, 3 hours, by herself to complete half the work. Then, together they both completed it in half an hour. Now, you must summarize your given data.

3 hours = Half of the work for Jen

30 Minutes = Jen and Maryna Together To Finish

If the two girls were working at the same pace, it would take them an hour and a half. However, it only took them 30 minutes to complete that. Now, we realize that Maryna works 3x faster than Jen does.

The question is asking, how long would it take her to FINISH the job.

So that means, we must divide 3 by Jen's time.

3 / 3 = 1

So, that means that it would only take Maryna 1 hour to complete the rest of the job.

Answer:

x=25

All triangles must equal 180

95+35=130

Answer:

15

Step-by-step explanation:

Answer: Look at picture

Answer:

a.) 0,51

b.) 51%

c.) 1/100

d.) 1%

e.) 3/20

f.) 0,15

Step-by-step explanation:

[tex]a.) \: \frac{51}{100} = 51 \div 100 = 0.51[/tex]

[tex]b.) \: \frac{51}{100} = 51\%[/tex]

[tex]c.) \: 0.01 = \frac{1}{100} [/tex]

[tex]d. )\: 0.01 = \frac{1}{100} = 1\%[/tex]

[tex]e.) \: 15\% = \frac{15}{100} = \frac{3}{20} [/tex]

[tex]f.) \: 15\% = \frac{15}{100} = 0.15[/tex]

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Answer: t<40

8p+<184

Step-by-step explanation:

The inequalities representing the given condition will be [tex]t\leq40[/tex] and [tex]8p+t\leq184[/tex].

Given information:

A baker makes apple tarts and apple pies each day.

The baker receives a shipment of 184 apples every day.

A tart requires 1 apple and a pie requires 8 apples.

Let t be the number of tarts made each day and p be the number of pies made each day.

The baker makes no more than 40 tarts per day. So, the condition can be written as,

[tex]t\leq40[/tex]

And the condition for each day apple delivery can be written as,

[tex]8p+t\leq184[/tex]

Therefore, the inequalities representing the given condition will be [tex]t\leq40[/tex] and [tex]8p+t\leq184[/tex].

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Use the old exp-log trick and properties of the logarithm to rewrite the limit as

[tex]\displaystyle \lim_{n\to\infty} \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)^{\frac xn}[/tex]

[tex]\displaystyle = \exp\left[\lim_{n\to\infty} \frac xn \ln \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac nk\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\left(\frac nk\right)^2\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac1{k/n}\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\frac1{\left(k/n\right)^2}\right)\right][/tex]

The first limit converges to 0, since n! asymptotically behaves like nⁿ, so ln(nⁿ/n!) → ln(1) = 0.

The two remaining sums converge to definite integrals:

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x + \frac1{\frac kn}\right) = \int_0^1 \ln\left(x + \frac1y\right) \, dy = \frac{(x+1) \ln(x+1)}x[/tex]

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x^2 + \left(\frac1{\frac kn}\right)^2\right) = \int_0^1 \ln\left(x^2 + \frac1{y^2}\right) \, dy = \frac{2\tan^{-1}(\sqrt x)}{\sqrt x} + \ln(1+x)[/tex]

It follows that

[tex]f(x) = \exp\left[x\left(0 + \dfrac{(x+1)\ln(x+1)}x - \dfrac{2\tan^{-1}(\sqrt x)}{\sqrt x} - \ln(x+1)\right)\right][/tex]

[tex]f(x) = \exp\left[\ln(x+1) - 2\sqrt x \tan^{-1}(\sqrt x)\right][/tex]

[tex]f(x) = (x+1) e^{-2\sqrt x \tan^{-1}(\sqrt x)}[/tex]

By computing f'(x) and f''(x), it's easy to show that f'(x) ≤ 0 and f''(x) ≥ 0 for all x > 0. So f(x) is decreasing and f'(x) is increasing, and

• (A) f(1/2) ≥ f(1) is true

• (B) f(1/3) ≤ f(2/3) is false

• (C) f'(2) ≤ 0 is true

• (D) f'(3)/f(3) ≥ f'(2)/f(2)   ⟺   f'(3)/f'(2) ≥ f(3)/f(2)   ⟺   (something larger than 1) ≥ (something smaller than 1) is true

Answer:

Question (a)

[tex]x^2-a^2-bx-ab=0[/tex]

Write in standard form [tex]ax^2+bx+c=0[/tex]:

[tex]\implies x^2-bx-(a^2+ab)=0[/tex]

Therefore,

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{b\pm\sqrt{(-b)^2+4(1)(a^2+ab)} }{2(1)}[/tex]

[tex]\implies x=\dfrac{b\pm\sqrt{b^2+4a^2+4ab} }{2}[/tex]

[tex]\implies x=\dfrac{b\pm\sqrt{(b+2a)^2} }{2}[/tex]

[tex]\implies x=\dfrac{b \pm (b+2a)} {2}[/tex]

[tex]\implies x=\dfrac{2b+2a} {2}=b+a[/tex]

[tex]\implies x=\dfrac{-2a} {2}=-a[/tex]

Question (b)

[tex]6x^2-15ax=2bx-5ab[/tex]

Write in standard form  [tex]ax^2+bx+c=0[/tex]:

[tex]\implies 6x^2-15ax-2bx+5ab=0[/tex]

[tex]\implies 6x^2-(15a+2b)x+5ab=0[/tex]

Therefore,

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{(-15a-2b)^2-4(6)(5ab)} }{2(6)}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{225a^2+60ab+4b^2-120ab} }{12}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{225a^2-60ab+4b^2} }{12}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{(15a-2b)^2} }{12}[/tex]

[tex]\implies x=\dfrac{(15a+2b) \pm (15a-2b) }{12}[/tex]

[tex]\implies x=\dfrac{30a }{12}=\dfrac52a[/tex]

[tex]\implies x=\dfrac{4b}{12}=\dfrac13b[/tex]

Question (c)

[tex]\dfrac{x^2}{x-1}=\dfrac{a^2}{2a-4}[/tex]

Cross multiply:

[tex]x^2(2a-4)=a^2(x-1)[/tex]

Write in standard form  [tex]ax^2+bx+c=0[/tex]:

[tex](2a-4)x^2-a^2x+a^2=0[/tex]

Therefore,

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{a^2\pm\sqrt{(-a^2)^2-4(2a-4)(a^2)} }{2(2a-4)}[/tex]

[tex]\implies x=\dfrac{a^2\pm\sqrt{a^4-8a^3+16a^2} }{4a-8}[/tex]

[tex]\implies x=\dfrac{a^2\pm\sqrt{a^2(a-4)^2} }{4a-8}[/tex]

[tex]\implies x=\dfrac{a^2\pm a(a-4)}{4a-8}[/tex]

[tex]\implies x=\dfrac{a^2\pm (a^2-4a)}{4a-8}[/tex]

[tex]\implies x=\dfrac{2a^2-4a}{4a-8}=\dfrac{2a(a-2)}{4(a-2)}=\dfrac12a \ \ \ (a\neq 2)[/tex]

[tex]\implies x=\dfrac{4a}{4a-8}=\dfrac{a}{a-2} \ \ \ \ (a\neq 2)[/tex]

Answer:

10

Step-by-step explanation:

25 times .60 then take that answer and subtract it from 25 or half of 25 which it 12.5 then take off another 10%

Answer:

5

Step-by-step explanation:

6 / 2 = 15 / l

3 = 15 / l

l = 15 / 3

l = 5

Answer:

three

Step-by-step explanation:

The lateral faces are any faces besides the base.

In summary, we will count how many sides there are in a triangular pyramid, and subtract one for the base.

-> A triangular pyramid has four sides / faces

-> A triangular pyramid has three lateral faces

Answer: 3

Step-by-step explanation: i did the quiz

Answer:

4x - 8 = 15

4x = 15 + 8

4x = 23 (divide both sides by 4 to get x)

4x/4 = 23/4

x = 5.75

Step-by-step explanation:

x = 5.75!!!!!

Answer:

I guess how many times out of x they will do something.

for example how many times out of 10 will bob ask out his crush vs Joe

Step-by-step explanation:

There isn't really an easy way to measure this.

The scale of measurement most reasonably applies to Confidence in Education would be the Ordinal scale.

The scale of measurement deals with how the variables are defined and categorized.

There are 4 types of scales of measurement;

1. Nominal scale

2. Ordinal scale

3. Interval scale

4. Ratio scale

We know that the Ordinale scale is used for variables in ranked order but the difference between them could not be determined.

Hence, The scale of measurement most reasonably applies to Confidence in Education would be the Ordinal scale.

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Answer:

729 : 15,625

Step-by-step explanation:

the ratio of their areas is 27² : 125² = 729 : 15,625

Answer:

The second option: Waxing crescent - first quarter; Waning gibbous - third quarter

Answer: A) 21,369

Step-by-step explanation: since 21,369 can be divided by 3 and = a whole number then it's divisible and also it's divided without a remainder / it's compatible so, 21,369/3= 7,123 which makes (A) your correct answer

Answer:

32 Bags.

Step-by-step explanation: