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3 years ago

What statement is true?

Mathematics

2 answers:

expeople1 [14]3 years ago

6 0

(A) because a right angle is 90 degrees and has a equal eqivalent to get to 100

Lana71 [14]3 years ago

6 0

Answer:

Option A is true

Step-by-step explanation:

Consider a right triangle ABC right angled at B.

Then we have A+C=90 since sum of all 3 angles =180

A and C are complementary

A has opposite side as BC and hypotenuse as AB

sin A = BC/AB

Consider angle B.

B has adjacent side as BC and hypotenuse as AB

So cos B = BC/AB

i.e. sin A = cos B

Hence we have option A to be true.

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Compare -3.6 and 1.2.

jasenka [17]

Answer:

-3.6 is a negative number compared to 1.2. 1.2 is -3 times less than -3.6. -3.6 is -3 times more than 1.2

Step-by-step explanation:

(can I get brainliest)

8 0

3 years ago

Read 2 more answers

Which is the graph of the linear inequality 2x – 3y < 12?

lora16 [44]

On a coordinate plane, a dashed straight line has a positive slope and goes through (0, negative 4) and (3, negative 2) and to the left of the line is shaded.

<h3>What are the characteristics of the graph of the inequality?</h3>

Inequality of a graph is represented with the greater then(<), less then(>) or with the other inequity signs. The inequality line on the graph is represented with the dotted lines.

Characteristics of the graph of the inequality-

Open circle-When the value of the variable is equal to the given number, then the graph of the inequality has an open circle.
Closed circle-When the value of the variable is not equal to the given number, then the graph of the inequality has a closed circle.
The ray will move to the right-When the, value of the variable is greater than the number, then the ray will move to the right.
The ray will move to the left-When the value of the variable is less than the number, then the ray will move to the left.

The linear inequality given in the problem is,

$2x - 3y < 12$

The graph of this line is attached below. Here, for the greater than sign, every thing on the left of the graph is shaded, and the line will be dotted because of inequality as shown in the attached image.

Hence, on a coordinate plane, a dashed straight line has a positive slope and goes through (0, negative 4) and (3, negative 2) and to the left of the line is shaded.

Learn more about the characteristics of graph of the inequality here;

brainly.com/question/62792

#SPJ1

6 0

2 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

Which is bigger 7/8 or 0.78

Paladinen [302]

7/8..........................................

8 0

3 years ago

Read 2 more answers

PLEASE HELP ITS DUE TODAY URGENT!

vlada-n [284]

Step-by-step explanation:

7. ∆ABC = ∆ILH by SSS

as, AB = IL , BC = LH , CA = HI

8. ∆DEF = ∆AMS by ASA

as , angle D = angle A, EF = MS , angle F = Angle S

9. ∆JKL = ∆HAT by SAS

as, JK = HA , KL = AT , angle L = angle T

10. ∆ABC = ∆KPG by ASA

as , CA = GK, Angle c = angle G and Angle B = angle P

11. ∆ABC = ∆YDE by ASA

as , angle A = angle Y, AB = YD , angle B = angle D

12. ∆MNO = ∆SAK by ASA

as , Angle M = angle S, NO = AK, angle O = angle K

hope this answer helps you dear...and may u have a great day ahead!

7 0

3 years ago