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3 years ago

All of the triangles above are _____ triangles.

Mathematics

2 answers:

Yuki888 [10]3 years ago

7 0

They are all Isosceles

V125BC [204]3 years ago

3 0

Answer:

All of the triangles above are isosceles triangles.

Step-by-step explanation:

From the figure, it can be seen that there are three triangles.

For first triangle, two of the sides of this triangle are equal in measure that is 3 units, therefore the given triangle is an isosceles triangle because in isosceles triangle, two of the three sides of given triangle are equal.

For second triangle, two of the sides of this triangle are equal in measure that is 33 units, therefore the given triangle is an isosceles triangle.

For third triangle, two of the sides of this triangle are equal in measure that is 4 units, therefore the given triangle is an isosceles triangle.

Thus, All of the triangles above are isosceles triangles.

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Find the equations for the lines through the point (a,

ozzi

The slope of the perpendicular line is -1/m.

it passes through the point (a,c) therefore:-

y - c = -1/m(x - a)

y = -1/m x + a/m + c answer

Parallel line:-

c = m(a) + b
b = c - ma

y = mx + c - ma Answer

4 0

3 years ago

Find the values of x and y

zhuklara [117]

You can use a tangent:

$tangent=\dfrac{opposite}{adjacent}$

We have opposite = 17 and adjacent = x.

$\tan30^o=\dfrac{\sqrt3}{3}$

substitute:

$\dfrac{17}{x}=\dfrac{\sqrt3}{3}$ cross multiply

$x\sqrt3=(3)(17)$ multiply both sides by √3

$x(\sqrt3)(\sqrt3)=51\sqrt3$

$3x=51\sqrt3$ divide both sides by 3

$x=17\sqrt3$

Use the Pythagorean theorem:

$y^2=(17\sqrt3)^2+17^2\\\\y^2=289(\sqrt3)^2+289\\\\y^2=289\cdot3+289\\\\y^2=867+289\\\\y^2=1156\to y=\sqrt{1156}\\\\y=34$

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Other method.

$30^o-60^o-90^o$ triangle.

The sides are in the ratio $1:2:\sqrt3\to17:y:x$

Therefore

$17:(2\cdot17):(17\sqrt3)\to17:34:17\sqrt3\to x=34,\ y=17\sqrt3$

4 0

3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago

If two cylinders are similar and the ratio between the altitude lengths is 2:3, what is the ratio of their volumes

hram777 [196]

Given:

ratio of altitude height = 2/3

Required:

ratio of volume

Solution:

Assuming that the only difference that the cylinders has is the height, we can solve for the ratio of the volume.

The volume of a cylinder is equal to πr²×height.

ratio of volume = πr²×2/πr²×3

We cancel the pi and the r² since we assume that the cylinders have the same radius.By cancelling, we are left with:

ratio of volume = 2/3

8 0

3 years ago

Grandma is making a

Svetlanka [38]

Answer:

the answer is 2304

Step-by-step explanation:

8×8=64

64×36=2304

4 0

3 years ago