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3 years ago

A recipe calls for 1 1/4 cups of sugar to make 20 cookies. How many cups of sugar will you need to make 14 cookies?

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

8 0

Probably 9/4 cups? Probably because it is lesser than what your asking?

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Are there less than 1 million, exactly 1 million, or greater than 1 million milligrams in 1 kilograms?

shusha [124]

Answer: There is exactly 1 million in 1 kilogram

Step-by-step explanation:

6 0

3 years ago

Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage r

Mazyrski [523]

Answer:

Decrease of 63.5%.

Step-by-step explanation:

Exponential function:

An exponential function has the following format:

$y = y(0)(1+r)^{x}$

In which y(0) is the initial value and r is the rate of change, as a decimal. If r is positive, it is a increase. If it is negative, it is a decrease.

In this question:

We have that:

$y = 3100(0.365)^x$

$1 - r = 0.365$

$r = 0.365 - 1 = -0.635$

So a decrease of 63.5%.

6 0

3 years ago

Read 2 more answers

8 x <img src="https://tex.z-dn.net/?f=10%5E%7B-3%7D" id="TexFormula1" title="10^{-3}" alt="10^{-3}" align="absmiddle" class="lat

schepotkina [342]

0.003- 0.000004= 0.002996

6 0

3 years ago

A container can hold 25 pounds of beans. Right now there are 12.5 pound of beans. What is the percentage of beans in the contain

aivan3 [116]

50% because 12.5lbs is half of 25lbs

6 0

3 years ago

Read 2 more answers

A flagpole is secured on opposite sides by two guy wires, each of which is 5 ft longer than the pole. the distance between the p

RideAnS [48]

The height of the flagpole to the nearest inch is 388 inches.

Explanation

According to the below diagram, AB is the flagpole which is secured on opposite sides by two guy wires, AB and AC.

Suppose, the length of the flagpole(AB) is $x$ ft.

As each of the guy wires is 5 ft longer than the pole, that means $AC= AD= (x+5)ft$

Also given that, the distance between the points where the wires are fixed to the ground is equal to the length of one guy wire. That means, $CD= (x+5)ft$

As $\triangle ABC$ and $\triangle ABD$ are congruent to each other, so $BC=BD= \frac{1}{2}CD= \frac{1}{2}(x+5) ft$

Now in right triangle $ABD$ , using Pythagorean theorem......

$AB^2+BD^2= AD^2\\ \\ x^2+[\frac{1}{2}(x+5)]^2= (x+5)^2\\ \\ x^2+ \frac{1}{4}(x+5)^2 = (x+5)^2\\ \\ 4x^2+(x+5)^2= 4(x+5)^2\\ \\ 4x^2= 3(x+5)^2\\ \\ 4x^2=3(x^2+10x+25)\\ \\ 4x^2-3x^2-30x-75=0\\ \\ x^2-30x-75=0$

Using quadratic formula, we will get.......

$x=\frac{-(-30)\pm \sqrt{(-30)^2-4(1)(-75)}}{2(1)}\\ \\ x= \frac{30\pm \sqrt{1200}}{2}\\ \\ x= 32.3205..., x=-2.3205...$

(Negative value is ignored as the height of the pole can't be in negative)

So, the height of the flagpole will be: $32.3205... ft =(32.3205...\times 12)inches = 387.846... \approx 388 inches$

4 0

3 years ago