All categories

3 years ago

25 POINTS!!!!! add. express your answer as a fraction. Do not include spaces in your answer 5 1/4+ 15.5

Mathematics

2 answers:

ICE Princess25 [194]3 years ago

8 0

Answer:

$20\frac{3}{4}$

Step-by-step explanation:

=> $5\frac{1}{4} + 15\frac{1}{2}$

=> $\frac{21}{4}+ \frac{31}{2}$

=> $\frac{21+62}{4}$

=> $\frac{83}{4}$

=> $20\frac{3}{4}$

ss7ja [257]3 years ago

5 0

Answer:

The answer is 83/4

Step-by-step explanation:

5 1/4+15.5

21/4+15.5

Taking LCM

21/4+15.5*4/1*4

21/4+62/4

83/4

You might be interested in

Melinda rod her bike 54/100 mile to the library. then she rode 4/10 mile to the store. how far did Melinda ride her bike in all?

Lady_Fox [76]

94/100 miles is the answer. I dont know what form you wanted the answer in but yeah.

3 0

3 years ago

I need help understanding this and help with the problems...

Pavel [41]

1. 15+37=?

15=10+5 (given on the diagram)
37=30+7 (given on the diagram)
add digits to digits 5+7=12=10+2
add tens to tens 10+30+10=50
add tens and digits 50+2=52.

2. 22+49=?

22=20+2 (given on the diagram)
49=40+9 (given on the diagram)
add digits to digits 2+9=11=10+1
add tens to tens 20+40+10=70
add tens and digits 70+1=71.

3. 38+26=?

38=30+8 (given on the diagram)
26=20+6 (given on the diagram)
add digits to digits 8+6=14=10+4
add tens to tens 30+20+10=60
add tens and digits 60+4=64.

Answer: 1) 52; 2) 71; 3) 64.

7 0

3 years ago

Read 2 more answers

What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?

natulia [17]

Answer:

<em>The endpoint coordinates for the mid-segment are: $(-2,-1)$ and $(1,0)$ </em>

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of $\triangle BCD$ are: $B(-3,1), C(3,3)$ and $D(-1,-3)$

Now, the endpoints of the mid-segment of $\triangle BCD$ which is parallel to $BC$ will be the mid-points of sides $BD$ and $CD$ .

<u>Formula for the coordinate of mid-point</u> : $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two endpoints.

So, the mid-point of $BD$ will be: $(\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)$

and the mid-point of $CD$ will be: $(\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)$

Thus, the endpoint coordinates for the mid-segment of $\triangle BCD$ that is parallel to $BC$ are $(-2,-1)$ and $(1,0)$

4 0

3 years ago

Read 2 more answers

What is the inverse of f(x)=x^-9

Leona [35]

$y = \frac{1}{x^9}\\yx^9 = 1\\x^9 = \frac{1}{y}\\x = \sqrt[9]{\frac{1}{y}}\\Therefore, \ f^{-1}(x)=\sqrt[9]{\frac{1}{x}}.$

3 0

3 years ago

(-3,2), (0, 3), (6,5)

Leni [432]

Answer: (3,8) and 2-,10) or the slope will be 3/2x

7 0

3 years ago