Which equation represents a quadratic function with a leading coefficient of 2 and a constant term of –3?

Answer:

31/24 menos 5/8 cuanto es 2/3

Step-by-step explanation:

=31/24- 5/8

=31-15/24

= 16/24

=2/3

Hence the final answer is 2/3

Answer:

12

Step-by-step explanation:

Step-by-step explanation:

if you draw an imaginary perpendicular line across the figure from the vertex which joins the line of 2 cm with the line that is making an angle of X then you can see that this figure is made up of two figures that is a triangle and a rectangle.

now from the angle given i.e X.

perpendicular= 5 cm

base = 14 -2= 12 cm

hypotenuse= ?

we know that,

h² = p²+b²

= 5²+12²=169

h= √169

h= 13

again,

cos X = b/h

= 12/13

Answer:

JMK

Step-by-step explanation:

Answer:

[tex]-\frac{1}{5}[/tex] is the slope of the line.

Step-by-step explanation:

(7 , 1) = (x1 , y1)

(-3 , 3) = (x2 , y2)

slope = y2 - y1/x2 - x1

=3 - 1/-3 - 7

=2/-10

=1/-5

=[tex]-\frac{1}{5}[/tex]

Answer:

1<1.35<2

Step-by-step explanation:

Answer:

1.35 is closer to 1 than 2

So the bag of nuts weighed about 1 pound.

Step-by-step explanation:

The decimal is closer to the whole number 1 when plotted on a number line, so it rounds to 1.

Answer:

Se requiere 34 meses para producir un interés de 8700 soles.

Step-by-step explanation:

Asumamos que el capital aumenta conforme a una tasa de interés compuesto, cuya fórmula se describe aquí abajo:

[tex]C = C_{o}\cdot \left(1 + \frac{r}{100} \right)^{t}[/tex] (1)

Donde:

[tex]C_{o}[/tex] - Capital inicial, en soles.

[tex]C[/tex] - Capital actual, en soles.

[tex]r[/tex] - Tasa de interés mensual, en [tex]\frac{1}{a}[/tex].

[tex]t[/tex] - Período de capitalización, en años.

Si sabemos que [tex]C_{o} = 42000[/tex], [tex]C = 50700[/tex] y [tex]r = 7\,\frac{\%}{a}[/tex], entonces el período de capitalización es:

[tex]C = C_{o}\cdot \left(1 + \frac{r}{100} \right)^{t}[/tex]

[tex]\frac{C}{C_{o}} = \left(1 + \frac{r}{100}\right)^{t}[/tex]

[tex]\ln \frac{C}{C_{o}} = t\cdot \ln \left(1 + \frac{r}{100}\right)[/tex]

[tex]t = \frac{\ln C - \ln C_{o}}{\ln \left(1+\frac{r}{100} \right)}[/tex]

[tex]t = \frac{\ln 50700 - \ln 42000}{\ln \left(1+\frac{7}{100} \right)}[/tex]

[tex]t = 2.782\,yr[/tex]

Es decir, se requiere 34 meses para producir un interés de 8700 soles.

Answer:

1

Step-by-step explanation:

that's a very vague question

Answer:

The answer is below

Step-by-step explanation:

The area of the wall is 84 ft².

The width of the wallpaper is 18 in.

1 ft. = 12 in

Hence; 18 in = 18 in * 1 ft. per 12 in = 1.5 ft

The length of the wallpaper is 33 ft.

Therefore, the area of the wallpaper = length * width = 1.5 ft. * 33 ft. = 49.5 ft². This means that each roll of wallpaper has an area of 49.5 ft²

Therefore, the minimum amount of rolls of wallpaper needed = area of wall / area of wallpaper

Amount of wallpaper = 84 ft² / 49.5 ft² = 1.7 rolls

Answer:

what do you mean for this question. What do you need help on?

Answer:

Part A

The volume of the triangular prism is 500 cm³

Part B

The total surface area of the prism is approximately 441.42 cm²

Step-by-step explanation:

The given details are;

The dimensions of the side length of the cube, s = 10 cm

The shape the cube was cut across to obtain = A prism

Part A

Whereby the prism obtained is a triangular prism, we have;

The cube can be cut in half to form a triangular prism

The volume of each triangular prism obtained = (1/2) × The volume of the cube

∴ The volume of the triangular prism = (1/2) × (10 cm)³ = 500 cm³

Part B

The height of the prism, h = 10 cm × sin(45°) = 5·√2 cm = (1/2) × The base width of the prism

The triangular cross sectional area of the prism, A₁ = 5·√2 × 5·√2 = 50

The square cross sectional area, A₂  = 10 × 10 = 100

The cross sectional area of the base, A₃ = 10·√2 × 10 = 100·√2

The total surface area of the prism, A = 2·A₁ + 2·A₂ + A₃

∴ A = 2×50 + 2×100 + 100·√2 = 300 + 100·√2 ≈ 441.42

The total surface area of the prism, A ≈ 441.42 cm²

Answer:

Step-by-step explanation:

x/7 - 13 ≥ -11

x/7 ≥ 2

x ≥ 14

m/-6 + 16 ≥ 19

m/-6 ≥ 3

m/6 ≤ -3

m ≤ -18

Question 23

Recall that a kite in geometry is any four sided figure where we have two different pairs of adjacent sides that are the same length.

We're told that PS = 4.5 and RS = 6. So this isn't an adjacent pair of congruent sides. But we can say that PS and PQ are one pair of adjacent congruent sides. That makes PQ to be 4.5 cm long.

Similarly, RS = RQ = 6 is the other pair of adjacent congruent sides. Notice that triangle PQR reflects over line PR to get triangle PSR.

See the diagram below.

Answer:

volume= 502.4cm

Step-by-step explanation:

see attachment for detailed solution

Answer:

Step-by-step explanation:

Formulae for Volume = Pi x r2 x h

Pi = 3.14 or 314/100 because the radius is not a product of 7

Radius = 4 cm

Height = 10 cm

314/100 x 4 x 4 x 10 = 512.4 cm2

= 512.4 cm2

Answer:

x= 52

Step-by-step explanation: you have a 90 degree angle the other side would be 90 also. you take 90 and subtract 38 to get the missing value?

Answer:

132'500

Step-by-step explanation:

It is like doing tax, you just needed to multiply your number by 0.6

Answer:

$ 7,500

Step-by-step explanation:

commission=6%

House price= $125,000

Commission=6/100 ×125,000

Commission= $ 7,500

Answer:

60 Degree,30 degree

Step-by-step explanation:

BD is bisector of ∠ABC

=> ∠ABD = ∠DBC

=> 3x = x + 20

=> 3x - x = 20

=> 2x = 20

=> x = 20/2  

=> x = 10

Now,

∠ABD = ∠DBC = x + 20 = 10 + 20 = 30 Degree

∠ABC = ∠ABD+∠DBC = 30+30 = 60 Degree

Answer:

(8,5):

8 is x-axis

5 is y-axis

it should be false but its been a while since i did that but the expressions should be multiplied together  

Answer: 52x+4

Step-by-step explanation: Do 16(2x-1)+4(5x+5) and get 52x+4.

Answer:

A: 42

Step-by-step explanation: 10.5x4=42

Answer:

your answer is A 42

Step-by-step explanation:

make me brainliest

Answer:

to do this lets first solve this so

5*5*5*5=625

5^-3=0.008

0.008^2=0.000064

0.000064*625=0.4

so all equations that are equal to 0.4 are right which is C

You can use the power rule for derivatives for each problem (though you could also use the product rule for the fourth curve).

y = x ² + 6x - 1   ==>   dy/dx = 2x + 6

y = x ² - 5x + 1   ==>   dy/dx = 2x - 5

y = 2 - 4x - x ²   ==>   dy/dx = -4 - 2x

y = (1 + x) (7 - x) = 7 + 6x - x ²   ==>   dy/dx = 6 - 2x

--

or, using the product rule,

dy/dx = (1 + x) (-1) + 1 (7 - x) = -1 - x + 7 - x = 6 - 2x

--

Now, stationary points occur where the derivative is zero. We have

2x + 6 = 0   ==>   x = -3

2x - 5 = 0   ==>   x = 5/2

-4 - 2x = 0   ==>   x = -2

6 - 2x = 0   ==>   x = 3

Answer:

Solution: {x = 2, y = -1/3}

Step-by-step explanation:

On observation, we see that the y-terms are equal and opposite in sign, therefore to solve the system, we add the two equations.

2x-3y = 5 ..........(1)

3x+3y = 5............(2)

(1) + (2)

2x - 3y + 3x + 3y = 5+5

5x = 10

x = 2 .......(3)

Substitute (3) in (2) and solve for x

3x+3y = 5 .....(2)

3(2) + 3y = 5

3y = 5-6 = -1

y = -1/3  .......(4)

assemble (3) and (4) to give the answer

Solution: {x = 2, y = -1/3}

Answer:

3  7/15

Step-by-step explanation:

1 4/5 + 2 2/3 - 16/15

9/5 + 8/3 - 16/15

27/15 + 40/15 - 16/15

52/15 = 3  7/15

Answer:

3 2/5

Step-by-step explanation:

STEP 1: Convert all fractions into improper fractions

9/5 + 8/3 - 16/15

Step 2: pass denominators

(9x3 + 8x5 - 16) / 15

=51/15

Step it into proper mixed number

3 2/5

Answer:

See Explanation

Step-by-step explanation:

You posted an incomplete question with little and unclear details.

I will answer this question with a more complete version (see attachment)

Given that:

[tex]Scores: 100\ 95\ 88\ 62\ 76\ 90\ 100\ 58\ 72\ 60\ 85\ 90\ 70\ 72\ 54\ 100\ 60\ 80\ 75\ 51[/tex]

Required

The fraction that passed the test

Rearrange the score in ascending order:

[tex]Scores: 51\ 54\ 58\ 60\ 60\ 62\ 70\ 72\ 72\ 75\ 76\ 80\ 85\ 88\ 90\ 90\ 95\ 100\ 100\ 100[/tex]

The total number of students is:

[tex]n =20[/tex]

Extract the students that passed (scored 70 and above)

[tex]Passed: 70\ 72\ 72\ 75\ 76\ 80\ 85\ 88\ 90\ 90\ 95\ 100\ 100\ 100[/tex]

Their numbers are:

[tex]Passed: 14[/tex]

So, the fraction of those that passed is:

[tex]Fraction = \frac{Passed}{n}[/tex]

[tex]Fraction = \frac{14}{20}[/tex]

Reduce fraction

[tex]Fraction = \frac{7}{10}[/tex]

Answer:

He made the following mistake, he assumed that polynomial [tex](x^{2}+1) = (x^{2}-1)[/tex], having for granted that [tex]x^{2}+1[/tex] has two real roots, instead of two complex roots.

Step-by-step explanation:

He made the following mistake, he assumed that polynomial [tex](x^{2}+1) = (x^{2}-1)[/tex], having for granted that [tex]x^{2}+1[/tex] has two real roots, instead of two complex roots. The true factorized form of the fourth grade polynomial is:

[tex]r(x) = (x^{2}+1)\cdot (x^{2}-9)[/tex]

[tex]r(x) = (x- i)\cdot (x+i)\cdot (x+3)\cdot (x-3)[/tex]

The lower the frequency the least common the average.

There are two frequencies that are 1, which would be the least common.

The answer is C. 0.320-0.329 and 0.360 -0.369

Answer:

option C

Step-by-step explanation:

option c is contains the two batting averages which have a frequency of 1, this means that they both only occurred once which is the least amount of times any of the batting averages has occurred.

Answer:

13[tex]\sqrt{22}[/tex] / 44

Step-by-step explanation:

cos A = 9/13

here adjacent = 9 and hypotenuse = 13  opposite = ?

using pythagoras theorem to find opposite

a^2 + b^2 = c^2

9^2 + b^2 = 13^2

81 + b^2 = 169

b^2 = 169 - 81

b^2 = 88

b = [tex]\sqrt{88}[/tex]

b =  [tex]2\sqrt{22}[/tex]

therefore opposite =  [tex]2\sqrt{22}[/tex]

cosec A = hypotenuse/opposite

= 13/[tex]2\sqrt{22}[/tex]

rationalizing the denominator

=13/ [tex]2\sqrt{22}[/tex]  *  [tex]2\sqrt{22}[/tex] /  [tex]2\sqrt{22}[/tex]

=13 *[tex]2\sqrt{22}[/tex] /( [tex]2\sqrt{22}[/tex] )^2

=26 [tex]\sqrt{22}[/tex] / 4*22

=26 [tex]\sqrt{22}[/tex] / 88

=13[tex]\sqrt{22}[/tex] / 44

Answer:

[tex]cosec A = \frac{13}{\sqrt{88}}[/tex]

   OR

[tex]cosec A = \frac{13 \sqrt{22}}{44}[/tex]

Step-by-step explanation:

Formulas used:

[tex]cos^2 A = 1 - sin^2 A\\\\cosec A = \frac{1}{sin A}[/tex]

Given :

  [tex]cos A = \frac{9}{13}[/tex]

Find cosec A

   [tex]sin ^2 A = 1 - cos^2 A[/tex]

              [tex]= 1 - (\frac{9}{13})^2\\\\= 1 - \frac{81}{169}\\\\=\frac{169 - 81}{169}\\\\=\frac{88}{169}[/tex]

 [tex]sin A = \sqrt{\frac{88}{169}} = \frac{\sqrt{88}}{13}[/tex]

Therefore,

         [tex]cosec A = \frac{1}{sin A} = \frac{1}{ \frac{\sqrt{88}}{13}} = \frac{13}{ \sqrt{88}}[/tex]

OR In a simplified form :

[tex]cosec A = \frac{13}{\sqrt{88}} \times \frac{\sqrt{88}}{\sqrt{88}} = \frac {13 \times \sqrt{4 \times 22}}{88} = \frac{13 \times 2 \sqrt{22}} {88} = \frac{13 \sqrt{22}}{44}[/tex]