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snow_tiger [21]
3 years ago
9

Molly's friend, Xavier, has 11/8 cups of strawberries. He needs 3/4 cup of strawberries to make a batch of tarts. How many batch

es can he make?
Mathematics
1 answer:
garik1379 [7]3 years ago
3 0

Answer:

1  5/6  batches

Step-by-step explanation:

to make 3/4 compatible with 11/8 you must multiply both the numerator and the denominator by 2 to get 6/8

he has 11/8 cups of strawberries

he needs 6/8 cups of strawberries

he can make 1 batch of tarts and he will have 5/8 cup strawberries left

divide 6/8 by 5/8 to get 5/6

he can make 1 5/6 batches of tarts

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Help plz helpppppppppppp plzzzzzzzzz ​
zzz [600]

it's 5 days, if you need an explanation feel free

5 0
3 years ago
Read 2 more answers
In 1998, Cathy's age is equal to the sum of the four digits in the year of her birthday, then how old was Cathy in 1998?
ycow [4]

Answer:

<em>Cathy was born in 1980 and she was 18 years old in 1998</em>

Step-by-step explanation:

<u>Equations</u>

This is a special type of equations where all the unknowns must be integers and limited to a range [0,9] because they are the digits of a number.

Let's say Cathy was born in the year x formed by the ordered digits abcd. A number expressed by its digits can be calculated as

x=1000a+100b+10c+d

In 1998, Cathy's age was

1998-(1000a+100b+10c+d)

And it must be equal to the sum of the four digits

1998-(1000a+100b+10c+d)=a+b+c+d

Rearranging

1998=1001a+101b+11c+2d

We are sure a=1, b=9 because Cathy's age is limited to having been born in the same century and millennium. Thus

1998=1001+909+11c+2d

Operating

88=11c+2d

If now we try some values for c we notice there is only one possible valid combination, since c and d must be integers in the range [0,9]

c=8, d=0

Thus, Cathy was born in 1980 and she was 18 years old in 1998. Note that 1+9+8+0=18

5 0
3 years ago
PLS HELP!!! FIRST ANSWER GETS BRAINLIEST!!!
olga55 [171]

Answer:C

Step-by-step explanation:11m is the radius and to find the diameter u multiply 11 x 2 which is radius x 2 which = 22

5 0
3 years ago
In the diagram shown of circle A, tangent MB is drawn along with chords BAC and BF . Secant MFE intersects BAC at G. It is known
mote1985 [20]

Answer:

  • arc BF = 76°
  • ∠M = 31°
  • ∠BGE = 121°
  • ∠MFB = 111°

Step-by-step explanation:

(a) ∠FBM is the complement of ∠FBC, so is ...

  ∠FBM = 90° -52° = 38°

The measure of arc BF is twice this angle, so is ...

  arc BF = 2∠FBM = 2(38°)

  arc BF = 76°

__

(b) ∠M is half the difference between the measures of arcs BE and BF, so is ...

  ∠M = (1/2)(138° -76°) = 62°/2

  ∠M = 31°

__

(c) arc FC is the supplement to arc BF, so has measure ...

  arc FC = 180° -arc BF = 180° -76° = 104°

∠BGE is half the sum of arcs BE and FC, so is ...

  ∠BGE = (1/2)(arc BE +arc FC) = (138° +104°)/2

  ∠BGE = 121°

__

(d) ∠MFB is the remaining angle in ∆MFB, so has measure ...

  ∠MFB = 180° -∠M -∠FBM = 180° -31° -38°

  ∠MFB = 111°

3 0
3 years ago
The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each
Fynjy0 [20]

Answer:

\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Step-by-step explanation:

Given equation of ellipsoids,

u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}

The vector normal to the given equation of ellipsoid will be given by

\vec{n}\ =\textrm{gradient of u}

            =\bigtriangledown u

           

=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})

           

=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}

           

=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}

Hence, the unit normal vector can be given by,

\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}

             =\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}

             

=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Hence, the unit vector normal to each point of the given ellipsoid surface is

\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

3 0
3 years ago
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