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4 years ago

5

Kim bought 2 pounds of green apples and 3.2 pounds of yellow apples from the market. Both apples were $1.65 per pound. How much

did Kim spend on apples?

1 answer:

Lesechka [4]4 years ago

7 0

Answer:

8.3325

Step-by-step explanation:

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HELP its urgent.............

irga5000 [103]

Answer:

first blank = 39

second blank = 28

Step-by-step explanation:

11 + __ + 7 + 28 = 85 = 39 + __ + 11+ 7

since,, four terms add upto form 85 out of which two terms (11 and 7) are common. so, the first blank will be filled with 39 and second blank with 28.

8 0

3 years ago

Read 2 more answers

The net of a prism is shown below.

Sliva [168]

Answer:

nothing is there

Step-by-step explanation:

7 0

3 years ago

What is the distance between the following two points?<br> (-5, -1) and (-8, -4)

liberstina [14]

The answer is -3 , -3

5 0

3 years ago

12/19*.............................= -12/19

melamori03 [73]

Answer:

$\frac{12}{19} \times \underline{\bold{-1}} = -\frac{12}{19}$

Step-by-step explanation:

$\frac{12}{19} \times ..... = -\frac{12}{19}$

= $\frac{\frac{12}{19}}{-\frac{12}{19}}$

= $\frac{12}{19} \times -\frac{19}{12}$

= $-\frac{228}{228}$

= $\bold{-1}$

So, the appropriate value to complete it is -1

#CMIIW ^^

3 0

4 years ago

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$

3 0

3 years ago