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2 years ago

Simplify the following expression by combining liker terms: 4x +9 -6x +6

Mathematics

2 answers:

Oxana [17]2 years ago

6 0

Answer:

its very simple you just need to collect like terms...

which is 4x-6x+6+9 which is

-2x+15.

This is my answer I hope it helps you.

mihalych1998 [28]2 years ago

6 0

Answer:

i think its -2x+15

Step-by-step explanation:

you do 4x-6x (-6 cuz it has a neg. in front) which gives you -2x then 9+6 and it gives you 15

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Elena va de compras con 400 euros se gasta 3/5 de esa cantidad ¿ cuanto le queda?

dexar [7]

240 euros

Explicación:

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1 year ago

Read 2 more answers

Need help with both of these please!! ASAP

Alik [6]

x = 1

Answer:

$m\angle XVW = 100\degree$

Step-by-step explanation:

In the first figure, FP is the bisector of $\angle DFE$

$\therefore m\angle 1 = m\angle 2\\\therefore 44x + 1 = 46x - 1\\\therefore 1 + 1 = 46x - 44x\\\therefore 2 = 2x\\\therefore \frac{2}{2} = x \\\huge \red {\boxed {\therefore x = 1}}$

In the second figure, VP is the bisector of $\angle XVW$

$\therefore m\angle 1 = m\angle 2\\\therefore 12x + 2 = 11x +6\\\therefore 12x - 11x = 6 - 2\\\therefore x = 4\\\because m\angle XVW = m\angle1 + m\angle 2\\\therefore m\angle XVW = 12x + 2 + 11x +6\\\therefore m\angle XVW = 23x + 8\\\therefore m\angle XVW = 23\times 4+ 8\\\therefore m\angle XVW = 92+ 8\\\huge \purple {\boxed {\therefore m\angle XVW = 100\degree}} \\$

8 0

3 years ago

Ihorangi travels 51 km to work and 51 km from work each day.

viktelen [127]

Answer:

Ihorangi uses 7.65 litres of petrol to and from work

Step-by-step explanation:

If the car Ihorangi drives consumes 7.5 L/100 km

Then for 1 km journey, the car will consume 7.5/100 L

= 0.075 L

If Ihorangi travels 51 km to work and 51 km from work

Therefore Ihorangi travel (51 + 51) km daily to and from work

= 102 km daily

The fuel he consumes daily is 102*0.075 L

= 7.65 L

7 0

3 years ago

Find the? inverse, if it? exists, for the given matrix.<br><br> [4 3]<br><br> [3 6]

True [87]

Answer:

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

$A A^{-1}=I$ where I is the identity matrix.

Consider, $A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]$

$\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}$

$\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0$

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$4\cdot \:6-3\cdot \:3=15$

$=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

4 0

3 years ago

An athlete goes on a diet. He loses 3 pounds each month for 8 months. He lost a total of 24 pounds. If he wants to lose 30 pound

sukhopar [10]

If he wants to loose 30 points total, subtract what he lost already, 24, from 30. He needs to lose 6 more pounds.

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2 years ago