Meena read 1/4 pages of a book containing 320 pages how many pages did she read?

Answers

Answer 1
Answer:

Step-by-step explanation:

Hope it will help you a lot.


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-l0 = -2+1/2 d solved for d

Answers

-10 =-2+ (1)/(2) d 

-10=-2+ (d)/(2) 

-10= (d)/(2) -2     Regroup \ terms 

-10+2= (d)/(2)     Add \ 2 \ to \ both \ sides 

-8= (d)/(2) 

-8*2=d      Multiply \ both \ sides \ by \ 2 

d=-16

Which of the following represents the zeros of f(x) = x3 − 12x2 + 41x − 42?7, −3, 2
7, −3, −2
7, 3, 2
7, 3, −2

Answers

Given the polynomial function f(x)=x^3-12x^2+41x-42.

The integer zeros should be divisors of free term -42.

The divisors of -42 are:

\pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42.

If some of these number is zero of f(x), then f takes 0 value at this point.

Check them:

f(1)=1^3-12\cdot 1^2 + 41\cdot 1-42=1-12+41-42=-12\neq 0;

f(-1)=(-1)^3-12\cdot (-1)^2 + 41\cdot (-1)-42=-1-12-41-42=-96\neq 0;

f(2)=2^3-12\cdot 2^2 + 41\cdot 2-42=8-48+82-42=0;

f(-2)=(-2)^3-12\cdot (-2)^2 + 41\cdot (-2)-42=-8-48-82-42=-180\neq 0;

f(3)=3^3-12\cdot 3^2 + 41\cdot 3-42=27-108+123-42=0;

f(-3)=(-3)^3-12\cdot (-3)^2 + 41\cdot (-3)-42=-27-108-123-42=-300\neq 0;

f(6)=6^3-12\cdot 6^2 + 41\cdot 6-42=216-432+246-42=-12\neq 0;

f(-6)=(-6)^3-12\cdot (-6)^2 + 41\cdot (-6)-42=-216-432-246-42=-936\neq 0;

f(7)=7^3-12\cdot 7^2 + 41\cdot 7-42=343-588+287-42=0.

Since the third degree polynomial function may have only 3 zeros, then you can end this process and state that zeros are 2, 3 and 7, because f(2)=0, f(3)=0 and f(7)=0.

Answer: correct choice is C

Answer: Zeroes are,

7, 3, 2

Step-by-step explanation:

Here, the given cubic equation,

f(x) = x^3 - 12x^2 + 41x - 42

Since, at x = 7,

f(7)=(7)^3-12* 7^2+41* 7 - 42 = 343 - 12* 49 +287 - 42 = 343 - 588 + 245=0

Thus, 7 is one of the zeroes of f(x),

⇒ x - 7 is a factor of f(x),

By the long division method ( shown below ),

We found that,

x^3 - 12x^2 + 41x - 42=(x-7)(x^2-5x+6)

=(x-7)(x^2-3x-2x+6) ( By middle term splitting )

=(x-7)(x(x-3)-2(x-3))

=(x-7)(x-3)(x-2)

For finding the zeroes, f(x) = 0,

(x-7)(x-3)(x-2)=0

⇒ x -7 =0 or x-3 =0 or x-2 =0

x = 7 or 3 or 2

ABCD is a square of side 3, and E and F are the mid points of sides AB and BC respectively. What is the area of the quadrilateral EBFD ?

Answers

Given : - Square ABCD with side 3.  E and F as midpoints.
To find : - area of EBFD

Solution : - We have, area of square ABCD = 3 x 3 = 9 units.

Thus, (ar)EBFD = ar ABCD - ar DAE - arDCF

arDAE = 1/2 x base x height

=1/2 x 1.5 x 3 ( AE is 1/2 of AB = 1.5, DA is altitude)
= 2.25

arDFC = 1/2 x base x height 
= 1/2 x 1.5 x 3 (FC is 1/2 of BC, DC is altitude)
= 2.25

Thus, (ar) EBFD = arABCD - arDAE - arDCF

= 9 - 2.25 - 2.25

= 4.5 units.

Thus, area of quad EBFD is 4.5 units.  

Plz answer this question...

Answers

Hello,

1) draw [AB]

2) draw mes (BAB')=78° (|AB|=|AB'| (distance))
draw the circle center B' passant by A.
3)C=middle [BB']

4) draw the half circle with [BB'] as diameter.

5) Intesection of cercle =>D

explain: any right triangle is inscribed in a semicircle

Rem: 78+90=168 and 360-168=192



What is the slope of the line shown below (2 2) (4 8) a. 3 b. 1/3 c. -1/3 d. -3

Answers

Answer:

Option A.3

Step-by-step explanation:

If its rise over run the fraction should be right 2 up 6 makeing a fraction of

6/2 which equals 3

The line has a slope of 3

angel has a total of 97 dimes and nickels. if the total value of the coins is $8.35, how many dimes does he have

Answers

dime=10 cents
nickles=5 cents

835=10d+5n
divide both sides by 5
167=2d+n

total number of coins is 97
97=d+n

we have

2d+n=167
d+n=97
multiply 2nd equation by -1 and add them

2d+n=167
-d-n=97 +
1d+0n=70

d=70

sub back

d+n=97
70+n=97
minus 70 both sides
n=27

70 dimes
He would have $1.35 left