All categories

3 years ago

In kite PQRS , TQ=3 cm and TP=4 cm. What is SP ? Enter your answer in the box.

Mathematics

2 answers:

Andru [333]3 years ago

5 0

Answer:5

Step-by-step explanation:

yawa3891 [41]3 years ago

4 0

Answer:

Kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent

Given: In kite PQRS

where PR and SQ are the diagonals of Kite respectively as shown in the figure given below.

Given: TQ = 3 cm and TP = 4 cm

Let PR is the main diagonal and SQ is the cross diagonal of kite PQRS as shown in figure, also let T is the intersection point of PQRS.

By Property of Kite, diagonal SQ bisects PR at perpendicular angle i.e, 90 degree.

i,e

Then, in right angle ΔQTP

$PQ = \sqrt{TQ^2+TP^2}$ [Using Pythagoras theorem]

Substitute the given values of TQ and TP we have;

$PQ = \sqrt{3^2+4^2} =\sqrt{9+16} =\sqrt{25} = 5 cm$

Also, PQ = SP [by definition of kite]

therefore, the side SP = 5 cm.

You might be interested in

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

3 years ago

Help please hurry i need done in 20 minutes

klemol [59]

Answer:

OKAY I WAS STRESSING FOR A QUICK MINUTE THINKING I WAS GETTING SOMETHING WRONG BUT THINK THE ANSWER IS CORRECT

So Cos0= 15/64 (SIMPLIFIED) or in decimal form 0.234375

Step-by-step explanation:

7/8² is 49/64 or (in decimal form) 0.765625

1 - 0.765625 = 0.234375

8 0

2 years ago

Factor out the coeffiecent:<br> 50 - 15<br> Please show work

quester [9]

just move the 15 the left and divide 5 and u should get c=3

7 0

3 years ago

Solve the equation for all real solutions in simplest form.<br> z^2 - 12z +9= -3

Mars2501 [29]

Answer:

Solving the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$ we get $\mathbf{z=10.9\: or\: z=1.1}$

Step-by-step explanation:

We need to solve the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$

First simplifying the equation:

$z^2 - 12z +9+3= -3+3\\z^2 - 12z +12= 0$

Now, we can solve the equation using quadratic formula:

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

we have a = 1, b=-12 and c=12

Putting values in formula and finding values of x

$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\z=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(12)}}{2(1)}\\z=\frac{12\pm\sqrt{144-48}}{2}\\z=\frac{12\pm\sqrt{96}}{2}\\z=\frac{12\pm9.8}{2}\\z=\frac{12+9.8}{2}\:or\:z=\frac{12-9.8}{2}\\z=10.9\:or\:z=1.1$

So, we get value of z: z=10.9 or z=1.1

Solving the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$ we get $\mathbf{z=10.9\: or\: z=1.1}$

3 0

3 years ago

Julie says ‘ I added three odd numbers and my answer was 50’ explainwhy Julie is incorrect

Dvinal [7]

Odd + odd = even + odd = odd

adding 3 odd numbers would make an odd answer

7 0

3 years ago

In ​ kite PQRS ​, TQ=3 cm and TP=4 cm. What is SP ? Enter your answer in the box.

In kite PQRS , TQ=3 cm and TP=4 cm. What is SP ? Enter your answer in the box.