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Tju [1.3M]
2 years ago
6

Let p: A number is greater than 25 Let q: A number is less than 35 If p ^ q is true, then what could the number be? Select two o

ptions. 24, 28, 32, 36, 40
Mathematics
1 answer:
luda_lava [24]2 years ago
8 0

Answer:

28,32

Step-by-step explanation:

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Convert 5 whole number and 7 over 8 into an improper fraction
Len [333]

Answer: 47/8

Step-by-step explanation: To write a mixed number as an improper fraction, first multiply the denominator by the whole number and then add the numerator. Finally, we will put our new number over our old denominator and we have our answer which is 47/8.

Therefore, 5\frac{7}{8} can be rewritten as the improper fraction 47/8.

7 0
2 years ago
A gym charges membership dues of $25 per month. Which equation represents the total cost, C, of belonging to the gym for m month
Anestetic [448]

Answer:

The answer is B.

Step-by-step explanation:

Given that the fixes amount is $25 per month. In order to find the total cost, C, you have to multiply $25 with the number of months.

C = 25m

6 0
3 years ago
Read 2 more answers
Modeling Exponential Growth and Decay In Exercises 119 and 120, Find the exponential function
Lubov Fominskaja [6]

Answer:

There is no exponential function passing through (1,1) and (5,5).

Step-by-step explanation:

We have the following exponential function

y = Ce^{kt}

The function passes through these two points:

(1,1): This means that when t = 1, y = 1

(5,5): This means that when t = 5, y = 5.

So

(1,1)

y = Ce^{kt}

Ce^{k} = 1

e^{k} = \frac{1}{C}

(5,5)

y = Ce^{kt}

5Ce^{k} = 1

Ce^{k} = \frac{1}{5}

From above, we have that:

e^{k} = \frac{1}{C}

C\frac{1}{C} = \frac{1}{5}

1 = \frac{1}{5}

1 cannot be equal to 1/5, so this is wrong.

This means that there is no exponential function passing through (1,1) and (5,5).

5 0
3 years ago
WILL MARK BRAINIEST<br> what type of gridline is labeled with a number?
ss7ja [257]

Answer:

I looked it up on Google and it said major and minor I don't know if this helps so I'm sorry if it doesn't

8 0
3 years ago
How to show these 2 problems are inverses
nataly862011 [7]
\bf \begin{cases}&#10;f(x)=\sqrt[3]{7x-2}\\\\&#10;g(x)=\cfrac{x^3+2}{7}&#10;\end{cases}\\\\&#10;-----------------------------\\\\&#10;now&#10;\\\\&#10;f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2}&#10;\\\\\\&#10;\sqrt[3]{x^3}\implies x\\\\&#10;-----------------------------\\\\&#10;or&#10;\\\\&#10;g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7}&#10;\\\\\\&#10;\cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x

thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other
8 0
2 years ago
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