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3 years ago

Can someone help me please

Mathematics

2 answers:

lesya692 [45]3 years ago

5 0

Answer:

a.(x)=7

b.(x)=14

Step-by-step explanation:

plz mark me as brainliest

andrey2020 [161]3 years ago

5 0

Answer:

For A x equals 7

For B x equals 14

Explanation:

Simply cross multiply

A. 21×2=42÷6=7

Or find the relationship

B. 150÷6=25

84÷6=14

Hope this helps, mark BRAINLIEST please...

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I have no idea!!! Help??

Savatey [412]

Answer:

x<5

Open circle at 5 going to the left

x >1

Open circle at 1 going to the right

Step-by-step explanation:

7x -19 < 16

Add 19 to each side

7x -19 +19< 16+19

7x< 35

Divide by 7

7x/7 < 35/7

x<5

Open circle at 5 going to the left

9+3x>12

Subtract 9 from each side

9-9 +3x >12-9

3x >3

3x/3 >3/3

x >1

Open circle at 1 going to the right

3 0

3 years ago

Explain why the quadratic relation y=2x^2+3x+5 would have no zeros

alex41 [277]

Complete the square to rewrite the quadratic:

2 x² + 3 x + 5 = 2 (x² + 3/2 x) + 5

... = 2 (x² + 3/2 x + 9/16 - 9/16) + 5

... = 2 (x² + 3/2 x + (3/4)²) + 5 - 9/8

... = 2 (x + 3/4)² + 31/8

Any real number squared becomes non-negative, so the quadratic expression has a minimum value of 31/8, which is greater than 0, and so there are no (real) x for which y = 0.

5 0

4 years ago

Solve the following equations. log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

4 years ago

PLEASE HELP FAST!!!!!!!!!!!

san4es73 [151]

Answer:

First expression 8y +4x +10

Second expression 7x-3y +6

If you want to combine both expressions: 5y +11x +16

Step-by-step explanation:

Like terms are the terms which have same variables and powers/roots.

Example: 7x and 2x are like terms because the variables are both "x". So we can combine in 9x

In an expression, only like terms can be combined. We combine like terms to shorten and simplify algebraic expressions, so we can work with them more easily. To combine like terms, we add the coefficients and keep the variables the same

10y + 3x + 10 + x -2y

10y-2y=8y 3x+x=4x

so 8y +4x +10

3x - y + 4x + 6 - 2y

3x+4x=7x -2y -y = -3y

so 7x-3y +6

4 0

3 years ago

If TU = 6 units, what must be true?

3241004551 [841]

Answer:

SU + UT = RT

Step-by-step explanation:

Given that TU = 6 units, RT = 12 units and RS = 24 units.

So,

For finding US, we will subtract the length of RT and TU from RS.

US = RS - RT - TU

US = 24 - 12 - 6

US = 6 units

Putting all the given values in the given conditions, we get the first option right, which is:

SU + UT = RT

$\rule[225]{225}{2}$

Hope this helped!

5 0

2 years ago