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mario62 [17]
3 years ago
5

Please please give answer for this question

Mathematics
1 answer:
Delvig [45]3 years ago
5 0
ANSWER: 25

EXPLANATION: First find the area of one single square by multiplying b(BASE) * h(HEIGHT). Repeat for other figure. Finally you will add both sums.
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6x+x^2 standard form​
Anna11 [10]

Answer:45

Step-by-step explanation:

8 0
3 years ago
Does x*x*x*x*x = 5x?<br> I'm not really sure so it would help.
hodyreva [135]

Answer:

Step-by-step explanation:

no ,x*x*x*x*x=x^5

x+x+x+x+x=5x

in multiplication those who have same base we add their power.

in addition those variables which have same base(or are like term) we add them.

6 0
3 years ago
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
3 years ago
In the past month, Rafael rented 4 video games and 1 DVD. The rental price for each video game was $2.30. The rental price for t
aliina [53]

Answer:

$12.50

Step-by-step explanation:

(2,30*4)+3.30

9.20+3.30=12.50

8 0
4 years ago
In order for an expression to be an equation, it must have​
wariber [46]

Answer:

An equal sign (=)

Step-by-step explanation:

i promise you it is right

6 0
3 years ago
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