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Liono4ka [1.6K]

2 years ago

10

A recipe that makes 4 servings calls for Two-thirds cup of flour. How much flour is required to make 20 servings? 6 cups 3 and o

ne-third cups two-thirds cups StartFraction 2 over 15 EndFraction cups

2 answers:

Vlad [161]2 years ago

6 0

Answer:

3 1/3 cups of flour

Step-by-step explanation:

20/4 = 5

20 servings is 5 times 4 servings, so you need 5 times the amount of ingredients.

5 * 2/3 = 10/3 = 3 1/3

Answer: 3 1/3 cups of flour

Verizon [17]2 years ago

4 0

Answer:

3 1/3 cups of flour is your answer.

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Louis started a table showing a multiplication patterns complete the table describe a pattern you see in the products

kvasek [131]

One pattern that you can see in a multiplication table is the perfect square numbers. It runs from the top left hand corner directly through the middle to the bottom right hand corner. A perfect square is a number that is multiplied by itself. The perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. They keep going forever on but those are the main ones from 1x1 to 12x12.

8 0

2 years ago

A car was travelling at a constant speed of 19 m/s, then accelerated for 10 seconds. The distance travelled, during this time, "

gregori [183]

Answer:

7.5 seconds.

Step-by-step explanation:

Substitute the distace 220m in the given equation:

d = 19t + t^2

200 = 19t + t^2

t^2 + 19t - 200 = 0

t = [-19 +/- sqrt(19^2 - 4*1*-200)] / 2

t = -19 +/- sqrt(1161) / 2

t = ( -19 +/- 34.07) / 2

t = 7.535 seconds

5 0

2 years ago

The rectangular playground in tim's school is three times as long as it is wide. the area of the playground is 75 square meters.

qwelly [4]

Let x be the width of the playground, then 3x is the length of the
<span>playground

х * 3х = 75
3x</span>² = 75
x² = 25
x = 5 m (width)
5*3=15 m (length)

Perimeter = 2(5+15) = 2*20 = 40 meters.

8 0

2 years ago

Consider the polynomial: StartFraction x Over 4 EndFraction – 2x5 + StartFraction x cubed Over 2 EndFraction + 1 Which polynomia

Phoenix [80]

Answer:

$- 2x^5+ 0x^4 + \frac{x^3}{2} +0x^2+\frac{x}{4} + 1$

Step-by-step explanation:

Given

$\frac{x}{4} - 2x^5 + \frac{x^3}{2} + 1$

Required

The standard form of the polynomial

The general form of a polynomial is

$ax^n + bx^{n-1} + cx^{n-2} +........+ k$

Where k is a constant and the terms are arranged from biggest to smallest exponents

We start by rearranging the given polynomial

$- 2x^5+ \frac{x^3}{2} +\frac{x}{4} + 1$

Given that the highest exponent of x is 5;

Let n = 5

Then we fix in the missing terms in terms of n

$- 2x^5+ 0x^{n-1} + \frac{x^3}{2} +0x^{n-3}+\frac{x}{4} + 1$

Substitute 5 for n

$- 2x^5+ 0x^{5-1} + \frac{x^3}{2} +0x^{5-3}+\frac{x}{4} + 1$

$- 2x^5+ 0x^{4} + \frac{x^3}{2} +0x^{2}+\frac{x}{4} + 1$

Hence, the standard form of the given polynomial is $- 2x^5+ 0x^4 + \frac{x^3}{2} +0x^2+\frac{x}{4} + 1$

3 0

2 years ago

For circle O, and m∠ABC = 55°. In the figure, ∠_____ and ∠____ have measures equal to 35°.

luda_lava [24]

Answer:

In the figure ∠ABO and ∠BCO have measures equal to 35°.

Step-by-step explanation:

<u><em>The complete question is</em></u>

For circle O, m CD=125° and m∠ABC = 55°

In the figure<____, (AOB, ABO, BOA) and <_____ (BCO, OBC,BOC) have measures equal to 35°

The picture in the attached figure

step 1

Find the measure of angle COB

we know that

$m\angle COB=arc\ CD$ ----> by central angle

we have

$arc\ CD=125^o$

therefore

$m\angle COB=125^o$

step 2

we know that

AB is a tangent to the circle O at point A

so

ABC and ABO are right triangles

In the right triangle ABC

Find the measure of angle BCA

Remember that

$m\angle BCA+m\angle\ ABC=90^o$ ---> by complementary angles in a right triangle

we have

$m\angle ABC=55^o$

substitute

$m\angle BCA+55^o=90^o$

$m\angle BCA=90^o-55^o=35^o\\$

step 3

In the triangle BCO

Find the measure of angle CBO

we know that

$m\angle CBO+m\angle COB+m\angle BCO=180^o$ ---> the sum of the interior angles in any triangle must be equal to 180 degrees

we have

$m\angle COB=125^o$

$m\angle BCO=m\angle BCA=35^o$ -----> have measure equal to 35 degrees

substitute

$m\angle CBO+125^o+35^o=180^o$

$m\angle CBO=180^o-160^o=20^o$

step 4

Find the measure of angle ABO

In the right triangle ABO

we know that

$m\angle ABC=m\angle CBO+m\angle ABO$ ----> by angle addition postulate

we have

$m\angle ABC=55^o$

$m\angle CBO=20^o$

substitute

$55^o=20^o+m\angle ABO$

$m\angle ABO=55^o-20^o=35^o$ ----> have measure equal to 35 degrees

therefore

In the figure ∠ABO and ∠BCO have measures equal to 35°.

3 0

2 years ago