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2 years ago

There are 7.55 grams of sugar in 5 servings of strawberries. How many grams of sugar are in a single serving of strawberries?

Mathematics

2 answers:

balandron [24]2 years ago

7 0

Answer:

1.51

Step-by-step explanation:

Divide 7.55 by 5 and it equals 1.51

Svetradugi [14.3K]2 years ago

6 0

The answer is 1.51. If you just divide 7.55 by 5 then you get your total of 1.51. To check your answer you can just take 1.51 and multiply it by 5 and you get 7.55.

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Check all the statements that are true:

sattari [20]

Answer:

The true among them are: B, C, E and F

Step-by-step explanation:

A is not true.

f(x) = x^x grows at of an order bigger than e^x

in fact f(x) = 4^x grows faster than e^x

B Is true... if p is growing more slowly than q then |x^p| < M|x^q| for all x> x0

C is true

D is not true.

f = O(h) and g = O(h)

fg = O(h^2)

E. Is true

F is true if |f(x) < M |g(x)| then |10 f(x)| < 10M |g(x)| and 10 M is a positive

7 0

3 years ago

Semmy [17]

Step-by-step explanation:

y=75+50(36 hours)=1875

give me brainliest

4 0

2 years ago

If the circumference is 3/2 pi what is the area

11111nata11111 [884]

Here is your answer

$\frac{9}{16} \pi \: {units}^{2}$

REASON:

Circumference(C)= 2πr

, where r is radius

Given

C= (3/2) π

So,

$2\pi \: r = \frac{3}{2}\pi$

$r = \frac{\frac{3}{2}\pi}{2\pi}$

$r = \frac{3}{4}$

Now,

$area = \pi {r}^{2}$

$= \pi \times { (\frac{3}{4}) }^{2}$

$= \frac{9}{16}\pi \: {unit}^{2}$

HOPE IT IS USEFUL

8 0

3 years ago

Consider the equation below -2(5x + 8) = 14 + 6x complete the statements below with the process used to achieve steps 1-4

Ad libitum [116K]

Step-by-step explanation:

1. Distribute the -2. multiply-2 with 5x and 8. it will be -10x-16=14+6x. For me, i make small numbers negative or positive than big ones. Add 16 to both sides. it will be -10x=30+6x. Subtract 6x. -10x-6x=-16x. Divide both sides by 16. x=30/16= 1 and 14/16

6 0

3 years ago

Which statement describes the inverse of m(x) = x^2 – 17x?

DochEvi [55]

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$ .

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$ .

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.

7 0

3 years ago