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2 years ago

Is it possible to draw an acute isosceles triangle with side lengths of 6 cm, 9 cm, and 12 cm and angles of 30°, 50°, and 100° ?

Mathematics

1 answer:

bagirrra123 [75]2 years ago

5 0

Answer:

Yes

Step-by-step explanation:

A full triangle is 180 degrees. 100 + 30 + 50 = 180.

I hope you get it right! :D

(i hope i get brainlist)

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Factor by grouping. a^2+2ab-24b^2

coldgirl [10]

You can rewrite that as:

a^2-4ab+6ab-24b^2 then factor 1st and 2nd pair of terms.

a(a-4b)+6b(a-4b) so you have

(a+6b)(a-4b)

3 0

3 years ago

Determine whether the following lines represented by the vector equations below intersect, are parallel, are skew, or are identi

KiRa [710]

Answer:

r(t) and s(t) are parallel.

Step-by-step explanation:

Given that :

the lines represented by the vector equations are:

r(t)=⟨1−t,3+2t,−3t⟩

s(t)=⟨2t,−3−4t,3+6t⟩

The objective is to determine if the following lines represented by the vector equations below intersect, are parallel, are skew, or are identical.

NOTE:

Two lines will be parallel if $\dfrac{x_1}{x_2}= \dfrac{y_1}{y_2}= \dfrac{z_1}{z_2}$

here;

$d_1 = (-1, \ 2, \ -3)$

Thus;

$r(t) = \dfrac{x-1}{-1} = \dfrac{y-3}{2}=\dfrac{z-0}{-3} = t$

$d_2 =(2, \ -4, \ +6)$

$s(t) = \dfrac{x-0}{2} = \dfrac{y+5}{-4}=\dfrac{z-3}{6} = t$

∴

$\dfrac{d_1}{d_2}= \dfrac{-1}{2} = \dfrac{2}{-4}= \dfrac{-3}{-6}$

Hence, we can conclude that r(t) and s(t) are parallel.

6 0

3 years ago

A circle with area 100π has a sector with a central angle of 2/5π radians.

Snowcat [4.5K]

Answer:

20π square units

Step-by-step explanation:

The sector area is found using the proportion ...

sector area / circle area = sector angle / whole circle angle

Multiplying by "circle area", we get ...

sector area = (circle area)( sector angle / whole circle angle)

= (100π)(2/5π)/(2π)

= (100π)(1/5)

sector area = 20π

3 0

3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago

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andrew-mc [135]

Answer:

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2 years ago