All categories

3 years ago

How do you solve this?

Mathematics

2 answers:

sesenic [268]3 years ago

5 0

Answer:

In general, we already know how to solve problems in a right triangle using the concept of trigonometry. Meanwhile, as we also know that the type of triangle is not only a right triangle, but there are isosceles, equilateral, or even random triangles. The question is how to solve the problems that exist in these triangles It is known that there is an arbitrary triangle with sides a, b, and c. The angle formed in front of side a is called angle α, the angle formed in front of side b is called angle β, and the angle formed in front of side c is called angle γ = 5

Step-by-step explanation:

x" + 5x + 6 = 0

(x + 2)(x + 3) = 0

x + 2 = 0

x = -2

x + 3 = 0

x = -3

#semogamembantu

nata0808 [166]3 years ago

5 0

Answer:

ac=10+5=15

AB=24

BC=5+(24-10)=5+14=19

perimeter=ac+AB+BC=15+24+19=58.the side in a tangant are equal.

You might be interested in

-3^2+5(-3)+9? the answer is not -15

lapo4ka [179]

Three is the answer!!!

7 0

3 years ago

Read 2 more answers

Find the area of each parallelogram. What is the relationship between the areas?

castortr0y [4]

Answer:

Area of parallelogram is given by:

$A = bh$

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12$ units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

7 0

3 years ago

What is 1/7 divided by what equals 14

Softa [21]

You would multiply 1/7 and 14 to get your number so your answer is 2

3 0

3 years ago

. Determine if the lines are parallel, perpendicular, or neither to the given line 24x + 12 y = 24

deff fn [24]

Answer:

a - neither b - perpendicular

Step-by-step explanation:

put them in desmos and look at the lines

5 0

3 years ago

Is it possible for an acute angle and obtuse angle to create complementary angles?

kenny6666 [7]

I’d say no because a complementary angle is 2 or more angles with a sum of 90 degrees [tip for differentiating supplementary and complementary angles: complementary (c) comes first in the alphabet (this is 90 degrees) and supplementary (s) comes later (this is 180 degrees. 180 comes later in counting] Anyway, an obtuse angle is more than 90 degrees which already can’t be a sum. (I’m not 100% sure but I hope this helps!)

7 0

3 years ago