Enter the answer to the problem below using the correct number of significant figures 125× 6.1

Answer:

28 4/9

Step-by-step explanation:

5 1/3 times 5 1/3

9514 1404 393

Answer:

  7

Step-by-step explanation:

If there is a common ratio, it can be found by finding the ratio of any two adjacent terms:

  42/6 = 7

The common ratio is 7.

Answer:

(3)(x + 4)(x-2)

Step-by-step explanation:

3x^2 + 6x – 24

3(x^2 + 2x - 8)

(3)(x + 4)(x-2)

Answer:

3(x - 2)(x + 4).

Step-by-step explanation:

3x^2 + 6x – 24

= 3(x^2 + 2x - 8)

= 3(x - 2)(x + 4).

Answer: (2,7) (-2,-9) (3,11) (-3,-13)

Step-by-step explanation:

(2)(2)+(2)(2)-1 = 7

(2) (-2)+(2)(-2)-1 = -9

(3)(2)+(2)(3)-1 = 11

(-3)(2)+(2)(-3)-1 = -13

Answer:

S = 35 + 20w

Step-by-step explanation:

Amount Samantha has in her account = $35

Additional savings per week = $20

Let

S =Totai savings in Samantha's account at the end of the week

w = number of weeks

The equation:

Totai savings in Samantha's account at the end of the week = Amount Samantha has in her account + (Additional savings per week * number of weeks)

S = 35 + 20w

Answer:

s=35+20w

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Polygon 3 and 5 are congruent cause they have the same length of side

The answer is B

Eplanation:

V = pi * r² * (h/3)

Then from here you just make h the subject of the formula:

h = (V * 3)/(pi * r²)

h = (393*3)/[3.14* (10/2)²]

h = 15.01910828 ft

which is rounded to 15 ft...B

:)

Answer:

27

Step-by-step explanation:

Answer:

Step-by-step explanation:

x² + 10x + 24 = 0

Discriminant = 10² - 4×1×24 = 4

The discriminant is positive, so there are two distinct, real solutions.

Answer:

The answer is C) 4; Two real solutions.

Step-by-step explanation:

x2 + 10x + 24 = 0 in the function  

x2 resembles a

10x resembles b

24 resembles c

so, let's convert.

b2-4ac

(10) ^2 - 4(1)(24)

100 - 96

= 4

Answer:

Step-by-step explanation:

Answer:

691 251/256

Step-by-step explanation:

So basically exponents is how much time you time the same number

First on the top is 2 exponent 4. that is 2x2x2x2=16

Second is 3 exponent 3. 3x3x3=27

Third is 9 exponent 7. 9x9x9x9x9x9x9=4782969

Fourth is 4 exponent 6. 4x4x4x4x4x4=4096.

All together is 16/27x4782969/4096=691 251/256

Answer:

Step-by-step explanation:

a) 3x^2

b) (512)^-2/3=2^n

1/64=2^n

n=-6

Answer:

Step-by-step explanation:

Answer:

x=8

Step-by-step explanation:

x^2+15^2=17^2, x=sqrt(64)=8

The equation has two real and equal roots for [tex]k = \frac{5}{6}[/tex]

In this question, we use the concept of the solution of a quadratic equation to solve it, considering that a quadratic equation in the format:

[tex]ax^2 + bx + c = 0[/tex]

has two equal solutions if [tex]\Delta = b^2 - 4ac[/tex] is 0.

------------------------------------

In this question:

The equation is:

[tex](2k+1)x^2 + 2x = 10x - 6[/tex]

Placing in the correct format:

[tex](2k+1)x^2 + 2x - 10x + 6 = 0[/tex]

[tex](2k+1)x^2 - 8x + 6 = 0[/tex]

Thus, the coefficients are: [tex]a = 2k + 1, b = -8, c = 6[/tex]

------------------------------------

Delta:

We want it to be positive, so:

[tex]\Delta = b^2 - 4ac[/tex]

[tex]\Delta = 0[/tex]

[tex]b^2 - 4ac = 0[/tex]

[tex](-8)^2 - 4(2k+1)(6) = 0[/tex]

[tex]64 - 48k - 24 = 0[/tex]

[tex]-48k + 40 = 0[/tex]

[tex]-48k = -40[/tex]

[tex]48k = 40[/tex]

[tex]k = \frac{40}{48}[/tex]

[tex]k = \frac{5}{6}[/tex]

The equation has two real and equal roots for [tex]k = \frac{5}{6}[/tex]

A similar question is found at https://brainly.com/question/12144265

Answer:

12

Step-by-step explanation:

4s multiples:

4,8,12,16,20,24

6s multiples:

6,12,18,24,30,36

lowest number that is a common multiple between both 4 and 6:

12

Answer: 2

Step-by-step explanation:

Answer:

[tex]\displaystyle d \approx 15.8768[/tex]

Step-by-step explanation:

We want to find the distance of d or AB.

From the right triangle with a 35° angle, we know that:

[tex]\displaystyle \tan 35^\circ = \frac{50}{PB}[/tex]

And from the right triangle with a 42° angle, we know that:

[tex]\displaystyle \tan 42^\circ = \frac{50}{PA}[/tex]

AB is PA subtracted from PB. Thus:

[tex]\displaystyle d = AB = PB - PA[/tex]

From the first two equations, solve for PB and PA:

[tex]\displaystyle \frac{1}{\tan 35^\circ } = \frac{PB}{50} \Rightarrow PB = \frac{50}{\tan 35^\circ}[/tex]

And:

[tex]\displaystyle \frac{1}{\tan 42^\circ } = \frac{PA}{50} \Rightarrow PA = \frac{50}{\tan 42^\circ}[/tex]

Therefore:

[tex]\displaystyle d = AB = \frac{50}{\tan 35^\circ} - \frac{50}{\tan 42^\circ}[/tex]

Using a calculator:

[tex]\displaystyle d= AB \approx 15.8768[/tex]

Answer:

false

Step-by-step explanation:

it would be in the first quadrant

Answer:

False

Step-by-step explanation:

The real part is positive and the imaginary part is positive which put is in the first quadrant

Answer:

y=3/x

Step-by-step explanation:

Answer:

1.92 m³

hopefully this answer can help you to answer the next question.

Answer:

[tex]\displaystyle [CQF]=5[/tex]

Step-by-step explanation:

Note that [tex][n][/tex] refers to the area of some polygon [tex]n[/tex].

Diagonal [tex]\overline{AC}[/tex] forms two triangles, [tex]\triangle ABC[/tex] and [tex]\triangle ADC[/tex]. Both of these triangles have an equal area, and since the area of parallelogram [tex]ABCD[/tex] is given as [tex]210[/tex], each triangle must have an area of [tex]105[/tex].

Furthermore, [tex]\triangle ADC[/tex] is broken up into two smaller triangles, [tex]\triangle ADF[/tex] and [tex]\triangle ACF[/tex]. We're given that [tex]\frac{DF}{FC}=2[/tex]. Since [tex]DF[/tex] and [tex]FC[/tex] represent bases of [tex]\triangle ADF[/tex] and [tex]\triangle ACF[/tex] respectively and both triangles extend to point [tex]A[/tex], both triangles must have the same height and hence the ratio of the areas of [tex]\triangle ADF[/tex] and [tex]\triangle ACF[/tex] must be [tex]2:1[/tex] (recall [tex]A=\frac{1}{2}bh[/tex]).

Therefore, the area of each of these triangles is:

[tex][ACF]+[ADF]=105,\\[][ACF]+2[ACF]=105,\\3[ACF]=105,\\[][ACF]=35 \implies [ADF]=70[/tex]

With the same concept, the ratio of the areas of [tex]\triangle AQE[/tex] and [tex]\triangle DQE[/tex] must be [tex]2:1[/tex] respectively, from [tex]\frac{AE}{ED}=2[/tex], and the ratio of the areas of [tex]\triangle DQF[/tex] and [tex]\triangle CQF[/tex] is also [tex]2:1[/tex], from [tex]\frac{DF}{FC}=2[/tex].

Let [tex][DQE]=y[/tex] and [tex][CQF]=x[/tex] (refer to the picture attached). We have the following system of equations:

[tex]\displaystyle \begin{cases}2y+y+2x=70,\\y+2x+x=35\end{cases}[/tex]

Combine like terms:

[tex]\displaystyle \begin{cases}3y+2x=70,\\y+3x=35\end{cases}[/tex]

Multiply the second equation by [tex]-3[/tex], then add both equations:

[tex]\displaystyle \begin{cases}3y+2x=70,\\-3y-9x=-105\end{cases}\\\\\rightarrow 3y-3y+2x-9x=70-105,\\-7x=-35,\\x=[CQF]=\frac{-35}{-7}=\boxed{5}[/tex]

Image of figure ABCD is missing and so i have attached it.

Answer:

D_new = (-1, - 12)

Step-by-step explanation:

From the figure attached, the current coordinates of D are; (-5, -2)

Now, we are told the figure undergoes a translation of (x,y) (x+6,y-10)

Thus, this means we add 6 to the x value and subtract 10 from the y-value.

Thus, new coordinate of D is;

> (-5 + 6, -2 - 10)

> (-1, - 12)

Answer:

1, -12

Step-by-step explanation:

D = -5, -2

|

-5 + 6 = 1

|

-10 and -2 is -12

1, -12

did it on edge, got it right.

This question is solved using proportions, and doing this, it is found that Evan gets 5 Chinese yuans for every Australian dollar.

-----------------------------------------

-----------------------------------------

75 dollars are worth 375 yuans. How much yuans is a dollar worth?

1 dollar - x yuans

75 dollars - 375 yuans

Applying cross multiplication:

[tex]75x = 375[/tex]

[tex]x = \frac{375}{75}[/tex]

[tex]x = 5[/tex]

Evan gets 5 Chinese yuans for every Australian dollar.

A similar question is given at https://brainly.com/question/23352001

Answer:

Given two equal chords. Since the line from the centre is always equal, thus the lines that bisect the two chords are also equal. X=5

Answer:

The angle θ = 312°.

Step-by-step explanation:

0° < θ < 360°

If cos θ = 2/3 and tan θ < 0

As cos θ is positive and tan θ is negative so the angle is in fourth quadrant.

cos θ = 2/3

θ = 312°

Answer:

4/5 or 0.8

Step-by-step explanation:

this problem description is not very precise. it leaves out the definition what corners or sides of JKLM correspond to corners and sides of ABCD.

I assume J and K correlate to A and B, and JK is a long side of JKLM.

so, we are going from JKLM to ABCD.

that means we are going from larger to smaller (as JK = 15 and therefore larger than AB = 12).

what is the scale factor to go from 15 to 12 ?

15 × x = 12

x = 12/15 = 4/5 or 0.8

Answer:

1) The total displacement vector is ((16 + √2)/2, -(6+11·√2)/2)

2) The number of miles they'd have to walk is approximately 13.856 miles

Step-by-step explanation:

1) The distance, direction, and location of the path of the walk the person takes,  are listed as follows;

Start location, (0, 0)

6 miles East walk to location, (6, 0)

6 miles Southeast to location, (6 + 3·√2, -3·√2)

3 miles South to location, (6 + 3·√2, -3·√2 - 3)

5 miles Southwest to location, (6 + 3·√2 - 2.5·√2, -3·√2 - 3 - 2.5·√2)

2 miles East to location, (6 + 3·√2 - 2.5·√2 + 2, -3·√2 - 3 - 2.5·√2)

(6 + 3·√2 - 2.5·√2 + 2, -3·√2 - 3 - 2.5·√2) = ((16 + √2)/2, -(6+11·√2)/2)

Therefore the destination coordinates is ((16 + √2)/2, -(6+11·√2)/2)

The total displacement vector, [tex]\underset{d}{\rightarrow}[/tex] = ((16 + √2)/2, -(6+11·√2)/2)

d =  (16 + √2)/2)·i - (6+11·√2)/2)·j

2) If the person walked straight home, the number of miles they'd have to walk, [tex]\left | \underset{d}{\rightarrow} \right |[/tex], is given as follows;

[tex]\left | \underset{d}{\rightarrow} \right | = \sqrt{\left(\dfrac{16 +\sqrt{2} }{2} \right)^2 + \left(-\dfrac{6 + 11 \cdot \sqrt{2} }{2} \right)^2 } = \sqrt{134 + 41 \cdot \sqrt{2} }[/tex]

Therefore;

If the person walked straight home, the number of miles they'd have to walk [tex]\left | \underset{d}{\rightarrow} \right | \approx 13.856 \ miles[/tex]