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2 years ago

Jack has 1 3/5 cups of yogurt to make smoothies. Each smoothie uses 2/5 How many smoothies can Jack make with the yogurt?

Mathematics

2 answers:

USPshnik [31]2 years ago

6 0

Answer:

Step-by-step explanation:

(1 3/5)/(2/5)

(8/5)/(2/5)

(8/5)(5/2)

40/10

Eddi Din [679]2 years ago

4 0

1 3/5 = 8/5
2/5x4 =8/5
It is 4 cups

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Which statement is true?

SVEN [57.7K]

Answer:

$y=log_{1} x$ is not a logarithmic function because the base is equal to 1.

Step-by-step explanation:

A logarithmic function is of the form: $y=log_{b}x$

For a logarithmic function to exist, we must fulfill certain conditions. The conditions are

i. The base, $b$ , is greater than 0. ${b>0)$

ii. The base, $b$ , is not equal to 1. $(b\ne 1)$

iii. $x$ must be greater than 0. $(x>0)$

Now, if we observe all the choices given, we conclude that the third choice is correct as the base there is 1. If base is equal to 1, then it violates the definition of logarithmic functions. So, the correct choice is:

$y=log_{1} x$ is not a logarithmic function because the base is equal to 1.

6 0

2 years ago

Read 2 more answers

the tallest tree in the forest is 150 feet from the water tower. unfortunatly in a lightning storm the tallest tree was struck a

gogolik [260]

The interior angles of a triangle add up to 180

180 - 97 - 34 = 49

150 / Sin 34 = b / Sin 49

multiply both sides by Sin 49

(Sin (49)* 150) / Sin 34 = b

b = 270.39 feet

3 0

3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$