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3 years ago

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Mathematics

2 answers:

gregori [183]3 years ago

7 0

Answer:

Third one.

x=6 y=12

Ok!

bekas [8.4K]3 years ago

7 0

Answer:

fourth option

Step-by-step explanation:

Given the 2 equations

y = 2x - 3 → (1)

x + y = 18 → (2)

Substitute y = 2x - 3 into (2)

x + 2x - 3 = 18

3x - 3 = 18 ( add 3 to both sides )

3x = 21 ( divide both sides by 3 )

x = 7

Substitute x = 7 into (1) for corresponding value of y

y = 2(7) - 3 = 14 - 3 = 11

Solution is x = 7, y = 11

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An observer (O) spots a plane flying at a 55° angle to his horizontal line of sight. If the plane is flying at an altitude of 21

xz_007 [3.2K]

Answer:

B. 25,636 feet

Step-by-step explanation:

sin(55) = h/x

h = 21,000

sin(55) = 21,000/x

x = 21,000/sin(55)

x = 25636.27 ≈ 25,636 feet

Hope this helped! :)

3 0

3 years ago

Read 2 more answers

Please help I will mark BRAINLIEST

Grace [21]

Here,

$1 \: batch \: requires \: 3 \frac{1}{2 } = 3.5 \: jars \: of \: pasta \: \\ sauce. \\ so \: 4 \frac{1}{2} = 4.5 \: batches \: requires \: 3.5 \times 4.5 \\ = 15.75 \: jars \: of \: \: pasta \: sauce.$

7 0

4 years ago

Express your answer in scientific notation 4.9x10^5-5.8x10^4

bonufazy [111]

Answer:

about 4.9x10^5

Step-by-step explanation:

(4.9x10^5-5.8x10^4)= 4899999.9994

3 0

3 years ago

In three baseball games over a weekend, 125,429 people came to watch. The next weekend, 86,353 people came to watch the games. H

topjm [15]

Just add the 125,429 and 86,353.

7 0

3 years ago

Read 2 more answers

How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (fourth-degree equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :

$t{ x }^{ 2 }+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

3 0

3 years ago

Read 2 more answers

Plsss help no spam......................​

Plsss help no spam......................