Ask question

Ask question

All categories

3 years ago

7

How many triangles exist with the given angle measures? 55°, 45°, 90° ?

2 answers:

tankabanditka [31]3 years ago

4 0

More than one triangle <span>exists with the given angle measures. So, this triangle is not equiangular, but has only 2 angles of equal measure.

Hope
this
helped</span>

Scilla [17]3 years ago

3 0

Many triangles............

You might be interested in

Carolyn paid a mechanic x dollars per hour for a 5 hour job plus $70 for parts. The total was $320. Which equation can be used t

Maslowich

70+5x=320
5x=250
x=50

The mechanic charged $50 per hour.

6 0

3 years ago

Read 2 more answers

Which of the following are incorrect expressions for slope? Check all that apply.. . A. Change in the dependent variable relativ

n200080 [17]

A. B. and C. are incorrect expressions for the slope.

m (slope) = rise/run = y2-y1/x2-x1 = change in the independent variable relative to the change in dependent variable

y is the independent variable and x is the dependent variable

7 0

3 years ago

Read 2 more answers

What is 3/m-2 = 7/m+2

kirill115 [55]

Answer:

3/m-2=7/m+2

3(m+2)=7(m-2)

3m+6=7m-14

3m-7m=-14-6

-4m=-20

m=20/4

m=5

6 0

3 years ago

use the pythagorean theorem to find the missing side length. 2 write sin, cosine, tangent of each angel.

Harlamova29_29 [7]

Answer:

1) $x=5\sqrt{3}$

2) $sin(a)=5\sqrt{3}/14$

$sin(b)=11/14$

$cos(a)=11/14$

$cos(b)=5\sqrt{3}/14$

$tan(a)=5\sqrt{3}/11$

$tan(b)=11/5\sqrt{3}$

Step-by-step explanation:

The Pythagorean Theorem is:

$a^{2}=b^{2}+c^{2}$

Where a is the hypotenuse and b and c are the legs.

The missing side lenght is one of the legs, then you must solve for one of them. Therefore, this is:

$x=\sqrt{(14yd)^{2}-(11yd)^{2}}=5\sqrt{3}yd$

MEASURE OF ANGLE:

Keep the identities on mind:

$sin\alpha=opposite/hypotenuse$

$cos\alpha=adjacent/hypotenuse$

$tan\alpha=opposite/adjacent$

Susbstitute values, then:

$sin(a)=5\sqrt{3}/14$

$sin(b)=11/14$

$cos(a)=11/14$

$cos(b)=5\sqrt{3}/14$

$tan(a)=5\sqrt{3}/11$

$tan(b)=11/5\sqrt{3}$

4 0

4 years ago

Read 2 more answers

Use the parabola tool to graph the quadratic function f(x)=x2−12x+27. Graph the parabola by first plotting its vertex and then p

alina1380 [7]

Answer:

vertex is (6 , -9)

points are (9,0) and (3,0)

Step-by-step explanation:

Given quadratic function $f(x)=x^2-12x+27$

We have to plot the given quadratic function.

Consider the Given quadratic function $f(x)=x^2-12x+27$

The general form of quadratic function is given $f(x)=a(x-h)^2+k$

Where, (h, k) is vertex , given $h =\frac{-b}{2a}$ and $k = f(h)$

Thus, for given quadratic function $f(x)=x^2-12x+27$

a = 1 , b= -12 , c = 27

Thus,

$h =\frac{-b}{2a}=\frac{12}{2}=6$

$k = f(h)$ that is f(12) = (6)^2 - 12× 6 +27 = 36 - 72 + 27 = - 9

Thus, given quadratic function $f(x)=x^2-12x+27$ in standard form is $f(x)=(x-6)^2-9$

Thus, vertex is (6 , -9)

For second point put $f(x)=0$ , we get,

$f(x)=(x-6)^2-9=0$

$\Rightarrow (x-6)^2-9=0$

$\Rightarrow (x-6)^2=9$

$\Rightarrow (x-6)=\pm 3$

$\Rightarrow x= 6\pm 3$

Thus, $\Rightarrow x=6+3=9$ and $\Rightarrow x=6-3=3$

thus, points are (9,0) and (3,0)

Graph is attached below.

5 0

4 years ago