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Gala2k [10]
3 years ago
15

Solve please. *Picture Included*

Mathematics
1 answer:
Juliette [100K]3 years ago
8 0
I think the answer would be 1.8
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-1.8(5) what’s the product
puteri [66]
The answer is -9 because you have to multiply the two numbers.
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3 years ago
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Which expression represents "4 less than the product of 2 and a number,<br> e"?
vlabodo [156]

Answer:

2 x e-4

Step-by-step explanation:

3 0
3 years ago
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This last one I need help on too
ExtremeBDS [4]

Answer:

x=\frac{3}{4}+i\frac{\sqrt{7}}{4},\:x=\frac{3}{4}-i\frac{\sqrt{7}}{4}

Step-by-step explanation:

simplify \frac{-3}{x-2} by putting the negative sign on the outside. \frac{2x}{x-1}-\frac{2x-5}{x^2-3x+2}=-\frac{3}{x-2}

find the LCM of the denominators. It is (x-1)(x-2). Multiply by the LCM:

\frac{2x}{x-1}\left(x-1\right)\left(x-2\right)-\frac{2x-5}{x^2-3x+2}\left(x-1\right)\left(x-2\right)=-\frac{3}{x-2}\left(x-1\right)\left(x-2\right)

Simplify:

2x\left(x-2\right)-\left(2x-5\right)=-3\left(x-1\right)

solve: x=\frac{3}{4}+i\frac{\sqrt{7}}{4},\:x=\frac{3}{4}-i\frac{\sqrt{7}}{4}

3 0
3 years ago
Solve the system of equations by substitution.<br> y = 5x - 13<br> 4x + 3y = 18<br> The solution is
Rudik [331]

Answer:

x=3 and y=2

Step-by-step explanation:

1.) 4x+3(5x-13)=18

expand into 4x+15x-39=18

18+39= 57

15x+4x= 19x

57/19= 3 so x=3

substitute 3 into y= 5x - 13

and you get 2 so y=2

3 0
3 years ago
What is the smallest 3-digit palindrome that is divisible by both 3 and 4?
kirill [66]

Answer:

252

Step-by-step explanation:

To be divisible by 3, it's digits have to add to a number that is a multiple of 3.

To be divisible by 4 its last 2 digits have to be divisible by 3.

So let's start with 1x1 which won't work because 1x1 is odd. so let's go to 2x2 and see what happens.

212 that's divisible by 4 but not 3

222 divisible by 3 but not 4

232 divisible by 4 but not 3

242 not divisible by either one.

252 I think this might be your answer

The digits add up to 9 which is a multiple of 3 and the last 2 digits are divisible by 4

4 0
2 years ago
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