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lukranit [14]
3 years ago
12

Ann puts 37 photos in one book and 24 photos in another book. how many photos does she use in all?​

Mathematics
2 answers:
Flura [38]3 years ago
6 0
Answer:
37 + 24 = 61
61 photos
Elodia [21]3 years ago
5 0

Answer:

61

Step-by-step explanation:

37+24=61

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Alexis has $3.25 in nickels and dimes. The number of dimes is 5 less than twice the number of nickels. Find the number of nickel
jek_recluse [69]

Answer:

15 nickels were there with Alexis

Step-by-step explanation:

Let there be  x nickels

given that

The number of dimes is 5 less than twice the number of nickels.

no of dimes in terms of x = 2*number of nickels - 5 = 2x - 5

we know that

value of 1 nickel = 5 cents

then

value of x nickel = 5*x cents = 5x cents

we know that

value of 1 dime = 10 cents

then

value of 2x-5 dime = 10(2x - 5)  cents = 20x - 50 cents

Total value of x nickel and 2x -5 dimes = 5x + 20x - 50 = 25x - 50

given that

Alexia has total of  $3.25 in nickel and dimes thus  25x - 50 must be equal to $3.25

$1 = 100 cents

$3.25 = 3.25 * 100 cents = 325 cents

Thus,

25x - 50 = 325

25x = 325 + 50 = 375

x = 375/25 = 15

Thus, no of nickels = x = 15  answer

5 0
3 years ago
Can someone help me with this??
Umnica [9.8K]

Answer:

divide by 2

Step-by-step explanation:

5 0
3 years ago
Need help with a equation
balandron [24]

26.6

RS is similar to NM, and TP is similar to OL.

If they're similar they have to be proportional. So:

\begin{gathered} \frac{NM}{RS}=\text{ }\frac{OL}{TP}\text{ } \\ \frac{7}{RS}=\frac{5}{19} \\ RS\text{ = }\frac{19\left(7\right)}{5} \\ RS=\text{ 133/5} \\ RS=26.6 \end{gathered}

7 0
1 year ago
A circle has a radius of 6 in. The circumscribed equilateral triangle will have an area of:
Ket [755]

Answer:

216

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
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