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kolezko [41]
3 years ago
15

How do I solve this?​

Mathematics
2 answers:
Brut [27]3 years ago
7 0

Answer:

x= -2

Step-by-step explanation:

antiseptic1488 [7]3 years ago
4 0

Answer:

x= -2

Step-by-step explanation:

y=x+4

y=2

2=x+4

-4+2=x

-2=x

x=-2

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7th grade math help me plzzz
murzikaleks [220]

Answer:

a. -7 + 3

b. draw 7 negatives + 7 positives

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
To prepare a chocolate cake, 2 1/4 cups of flour were used. How many grams of flour was used if each cup of flour weighs 120 gra
Marianna [84]
120/4 = 30
120 x 2 = 240
240 + 30 = 270
270 grams of flour were used
4 0
3 years ago
Given the matrices: 1 2 A= 1 -1 2 1 1 B= 3 4 Calculate AB: C11 C12 [2.1] х 1 2 3 4 C21 C22 C11 = C12 = -2 C22 - C215 DONE​
eduard

Answer:

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

Step-by-step explanation:

Given the matrices

A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}

B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}

Calculating AB:

\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}

Multiply the rows of the first matrix by the columns of the second matrix

                                   =\begin{pmatrix}1\cdot \:1+\left(-1\right)\cdot \:3&1\cdot \:2+\left(-1\right)\cdot \:4\\ 2\cdot \:1+1\cdot \:3&2\cdot \:2+1\cdot \:4\end{pmatrix}

                                   =\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}

Hence,

\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}

Therefore,

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

5 0
3 years ago
Read 2 more answers
Mum bought a litre of orange juice. She filled 5 glasses and had 1/5 of a litre left. How much was in each glass?
avanturin [10]
1/5 of a litre is 0.2 litres
1 litre minus 0.2 litres is 0.8 litres so
0.8 divided by 5 (glasses) is 0.16 litres in each glass
7 0
3 years ago
How to factor this expression <br><br>x^3 - 4x^2 + 4x - 16
Natalija [7]
X²(x - 4) +4 (x - 4)

(x² + 4) (x - 4)

First find the common terms that can enter into both x³ and 4x² then write its down in this case it’s x² that can enter x³ leaving only x _since x³/x² = subtract of the indices. x² will also enter 4x² leaving only four hence you having x² (x - 4)
then do the same for the next pair of terms giving you 4 that can enter into both 4 and 16
Leaving you with +4 (x - 4)

Now you can put the common terms together like so (x² + 4) and choose get one of the other two which are the same= (x - 4)
= (x² + 4) (x - 4)
8 0
3 years ago
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