5. The points (4, 1) and (x,-6) lie on the same line. If the slope of the line is 1, what is the value of

Three times one number added to five times another number is 54. The second number is two less than the first. Use a system of e

worty [1.4K]

Answer:

8 and 6

Step-by-step explanation:

If you use x for the first number, and y for the second, you get equations of 3x+5y=54 and x-2=y.

You can substitute the second equation into the first one to solve it. This gives 3x+5(x-2)=54.

The brackets can be expanded to 3x+5x-10=54. Collecting like terms makes it 8x-10=54.

Next, we add 10 to both sides. This gives 8x=64.

From there, we isolate the x by dividing both sides by 8, giving x=8.

To work out the second number, we can sub 8 for x in either equation (I'm using the second one as it's simpler).

This comes to y=8-2, therefore y=6.

**This content is simultaneous equations, which you may wish to revise. I'm always happy to help!

3 0

3 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

Which is the better deal, 139.99 with 20 percent in discountor 139.99 with $20 in discount

Leno4ka [110]

20% discount is definitely the $ saver!

3 0

3 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST !!

AveGali [126]

Answer:

Angles and XWY and STR

Step-by-step explanation:

These two angels are alternate exterior angels because:

- They are on different sides of the transversal (the line intersecting the two parallel lines)

- They have the same angle measure

- They are exterior, meaning they are on the outside, not inside (interior)

Hope this helps!!! :)

8 0

2 years ago

Simplify the square root of 27 over 12

Hunter-Best [27]

$\sqrt{ \frac{27}{12} }$

${ \frac{ \sqrt{27} {} }{ \sqrt{12}}$

$\frac{ \sqrt{9 * 3} }{ \sqrt{4 * 3} }$

$\frac{ \sqrt{9} \sqrt{3} }{ \sqrt{4} \sqrt{3} }$

$\frac{3 \sqrt{3} }{2 \sqrt{3} }$

$\frac{3}{2}$

$1 \frac{1}{2}$

5 0

3 years ago