Identify the quantities that are equivalent to 250 meters.

Ratio Conversion Table
kilometer (km) : meter (m) 1 : 1,000
meter (m) : centimeter (cm) 1 : 100
centimeter (cm) : millimeter (mm) 1 : 10

Answer:

Between 50 and 60

Step-by-step explanation:

4,346/82 is 53 which is between 50 and 60.

Hope this helps!

Answer:

1) Isosceles

2) Acute

3) Right angled

4( Obtuse

5) Equilateral

Answer:

the answer is 8

Step-by-step explanation:

Total amount Ted wishes to spend this month is $5000.

Mahogany costs $450 per block, and he has already bought 7 blocks of spruce at $200 each.

So, money spent on spruce =  dollars

Money left after buying spruce =  dollars

Now as $450 will be spent on buying 1 block of mahogany.

So, $3600 will be spent

and then your answer will be 8

Answer:

I don't think it's possible.

Answer:

"D" is the correct answer

Step-by-step explanation:

im struggling with the same one

f(0) = 56

f(2) = 42

f(-2) = 70

f(x+1) = 7|x-7|

f(x²+2) = 7|x²-6|

Answered by GAUTHMATH

Answer:

a)  b^5/12

Step-by-step explanation:

b^2/3 ÷ b^1/4    [if bases are the same then subtract the exponents]

2/3 - 1/4 = 5/12

Answer:

Volume - 582.8848178 cm cubed

Step-by-step explanation:

Volume - to fine the volume start with the cone...

The formula is 1/3 pi r^2h

so for this problem it would be

1/3 * 3.14 * 26.01 * 11.2

so your answer would be...

305.061213 cm cubed

Your next step is to find the volume of the dome. to do that the formula is (4/3 pi r^3)/2

so for this problem it would be

(4/3 * 3.14 * 132.651)/2

so your answer would be...

277.8236048 cm cubed

Finally you add together those 2 volumes and get 582.8848178 cm cubed as your final answer.

Answer:

b=2

Step-by-step explanation:

Step-by-step explanation:

[tex]\dfrac{dy}{dx} = \dfrac{2x}{y(x^2 + 1)}[/tex]

Rearranging the terms, we get

[tex]ydy = \dfrac{2xdx}{x^2 + 1}[/tex]

We then integrate the expression above to get

[tex]\displaystyle \int ydy = \int \dfrac{2xdx}{x^2 + 1}[/tex]

[tex]\displaystyle \frac{1}{2}y^2 = \ln |x^2 +1| + k[/tex]

or

[tex]y = \sqrt{2\ln |x^2 + 1|} + k[/tex]

where I is the constant of integration.

Answer:

[tex]b_0 = 16.471[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{ccccc}x & {15} & {12} & {10} & {7} \ \\ y & {5} & {7} & {9} & {11} \ \end{array}[/tex]

Required

The least square estimate [tex]b_0[/tex]

Calculate the mean of x

[tex]\bar x = \frac{\sum x}{n}[/tex]

[tex]\bar x = \frac{15+12+10+7}{4} =\frac{44}{4} = 11[/tex]

Calculate the mean of y

[tex]\bar y = \frac{\sum y}{n}[/tex]

[tex]\bar y = \frac{5+7+9+11}{4} =\frac{32}{4} = 8[/tex]

Calculate [tex]\sum(x - \bar x) * (y - \bar y)[/tex]

[tex]\sum(x - \bar x) = (15 - 11) * (5 - 8)+ (12 - 11) * (7 - 8) + (10 - 11) * (9 - 8)+ (7 - 11) * (11 - 8)[/tex]

[tex]\sum(x - \bar x) = -26[/tex]

Calculate [tex]\sum(x - \bar x)^2[/tex]

[tex]\sum(x - \bar x)^2 = (15 - 11)^2 + (12 - 11)^2 + (10 - 11)^2 + (7 - 11)^2[/tex]

[tex]\sum(x - \bar x)^2 = 34[/tex]

So:

[tex]b = \frac{\sum(x - \bar x) * (y - \bar y)}{\sum(x - \bar x)^2}[/tex]

[tex]b = \frac{-26}{34}[/tex]

[tex]b_0 = y - bx[/tex]

[tex]b_0 = 5 - \frac{-26}{34}*15[/tex]

[tex]b_0 = 5 + \frac{26*15}{34}[/tex]

[tex]b_0 = 5 + \frac{390}{34}[/tex]

Take LCM

[tex]b_0 = \frac{34*5+ 390}{34}[/tex]

[tex]b_0 = \frac{560}{34}[/tex]

[tex]b_0 = 16.471[/tex]

Hello,

Answer A

[tex]dom (B(n)) =\{0,1,2,3\} =\{ z\ in \ \mathbb{Z} \ |\ 0 \leq z \leq 4\}[/tex]

9514 1404 393

Answer:

  y = -4/7x +58/7

Step-by-step explanation:

The slope of the given line segment is ...

  m = (y2 -y1)/(x2 -x1)

  m = (17 -3)/(1 -(-7)) = 14/8 = 7/4

Then the slope of the perpendicular line is ...

  -1/m = -4/7 . . . . . slope of the perpendicular bisector.

__

The midpoint of the given line segment is ...

  M = 1/2(x1 +x2, y1 +y2)

  M = (1/2)(-7 +1, 3 +17) = 1/2(-6, 20) = (-3, 10)

__

The y-intercept of the bisector can be found from ...

  b = y -mx

  b = 10 -(-4/7)(-3) = 10 -12/7 = 58/7

Then the slope-intercept form equation for the perpendicular bisector is ...

  y = mx +b

  y = -4/7x +58/7

Let f(x) be the sum of the geometric series,

[tex]f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]

for |x| < 1. Then taking the derivative gives the desired sum,

[tex]f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}[/tex]

Answer:

a) false

b) true

c) true

d) true

e) false

Step-by-step explanation:

In a statement of the type:

p ∧ q

where ∧ means "and"

The statement is true only if both p and q are true

the statement is false if p, q, or both, are false.

and in the case of:

p ∨ q

where ∨ means "or"

The statement is true if at least one of p or q (or both) are true.

The statement is false if both are false.

Now that we know that, let's solve the problem:

a) "5 is an odd number and 3 is a negative number."

Here we have:

p = 5 is an odd number

We know that this is true

q = 3 is a negative number

This is false.

then the complete statement is false.

b) "5 is an odd number or 3 is a negative number."

here we have:

p = 5 is an odd number.

this is true

q = 3 is a negative number

because in this case we have an "or", with only p being true, the whole statement is true.

c) "8 is an odd number or 4 is not an odd number."

p = 8 is an odd number  (this is false)

q = 4 is not an odd number  (this is true, 4 is a even number)

Again, we have an "or", so we need only one true proposition, then the statement is true.

d) "6 is an even number and 7 is odd or negative."

p = 6 is an even number  (true)

q = 7 is odd or negative  (notice that we have an or, and 7 is odd is true, so this proposition is true)

Then both propositions are true, then the statement is true.

e) "It is not true that either 7 is an odd number or 8 is an even number (or both)."

This is most complex, this will be true if at least one of the propositions is false.

but:

7 is an odd number is true

8 is an even number is true.

Then both statements are true, which means that the statement is false.

9514 1404 393

Answer:

  2 -i

Step-by-step explanation:

The two diagonals have the same center point. Here, it is ...

  (1 +2) +(-1 -2i) = 0 +0i

The fourth point is found by reflecting the 2nd point across the origin:

  p4 = -p2 = -(-2 +i) = 2 -i . . . . . fourth corner of the square

Answer:

8. 36

Step-by-step explanation:

8.

The diameter of the square is x=[tex]\sqrt{6^{2} +6^{2} }\\[/tex] = [tex]\sqrt{72}[/tex] = 6 [tex]\sqrt{2}[/tex]

The diameter of the square is the diameter of the circle, therefore the radius of the circle is r = 6 [tex]\sqrt{2}[/tex]/2 = 3[tex]\sqrt{2}[/tex]

The area of the shaded region, which is a circle is A= π[tex]r^{2}[/tex] =  π[tex](3\sqrt{2} )^{2}[/tex] = 36π

Answer:

The numerical values of the circumference and area of the circle are equal.

Answer:

- 3.027

Step-by-step explanation:

First price = 10000 ; second price = 700

Number of tickets sold = 11000

Ticket cost = $4

Probability that a ticket wins grand price = 1 / 11000

Probability that a ticket wins second price = 1 / 11000

X ____ 10000 _____ 700

P(x) ___ 1 / 11000 ___ 1/11000

Expected winning for a ticket buyer :

E(X) = Σx*p(x)

E(X) = (1/11000 * 10000) + (1/11000 * 700) - ticket cost

E(X) = 0.9090909 + 0.0636363 - 4

E(X) = - 3.0272728

E(X) = - 3.027

Answer:

9:2

Step-by-step explanation:

Answer:

B.

Step-by-step explanation:

[tex] \frac{4 + 4i}{5 + 4i} [/tex]

[tex] \frac{4 + 4i}{5 + 4i} \times \frac{5 - 4i}{5-4i} [/tex]

[tex] \frac{(4 + 4i)(5 - 4i)}{(5 + 4i)(5 - 4i)} [/tex]

[tex] \frac{20-16i+20i-16i^2}{(5) {}^{2} - (4i) {}^{2} } [/tex]

(combining like terms)

[tex] \frac{20+(-16i+20i)-(-16)}{25-(-16)} [/tex]

[tex] \frac{(20+16)+4i}{25+16} [/tex]

[tex] \frac{36+4i}{41} [/tex]

distributing the denominator

[tex] \frac{36}{41} + \frac{4}{41}i [/tex]

That is, option B.

Answer:

[tex]\frac{20}{40} \ and \ \frac{4}{8} \ is \ equivalent[/tex]

Step-by-step explanation:

1.

[tex]\frac{2}{3} \ and \ \frac{12}{9} \\\\\frac{2}{3} \ and \ \frac{4}{3}\\\\Not \ equivalent[/tex]

2.

[tex]\frac{20}{40} \ and \ \frac{45}{55}\\\\\frac{1}{2} \ and \ \frac{9}{11}\\\\Not\ equivalent[/tex]

3.

[tex]\frac{20}{40} \ and \ \frac{4}{8}\\\\\frac{1}{2} \ and \ \frac{1}{2} \\\\Equivalent[/tex]

4.

[tex]\frac{5}{5} \ and \ \frac{25}{50} \\\\\frac{1}{1} \ and \ \frac{1}{2} \\\\not \ equivalent[/tex]

Answer:

.0421

about 4.21%

Step-by-step explanation:

[tex]\frac{{8\choose3}}{{21\choose3}}=\frac{56}{1330}=.042105263[/tex]

Answer:

Total no. of students =175

percent of students athletes who attend a preseason meeting =84

No. of students who attend the meeting

= 84% of 175

=147

Answer: 13/7 or as a decimal 1.857142857

How did i get the answer:

Step 1: Simplify both sides of the equation.

so 1/2 of 10 is 5, 1/2 of 16 is 8

-3/5 of 15 is -9 and -3/5 of -35 is POSITIVE 21

all together should look like 5x+8+−13=−9x+21

(now we have to combine like terms)

8+ -13= -5

5x -5 = -9x+21

Step 2: Add 9x to both sides

5x + 9x= 14x

14x -5 = 21

Step 3: Add 5 to both sides.

21+5= 26

14x=26

Step 4: Divide both sides by 14.

26/14= 1.85714286 or 13/7