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Leona [35]
3 years ago
14

Please help!!! Its an emergency

Mathematics
2 answers:
Sliva [168]3 years ago
7 0

Answer:

x=10

Step-by-step explanation:

The middle triangle has d as the hypotenuse and is an isosceles right triangle. The sides of such a triangle are in the ratio of y, y, and \sqrt{2}y.

For this triangle

\sqrt{2}y=5\sqrt{2}

so y=5.

Tthe triangle on the right is a 30, 60, 90 triangle with sides in the ratio of

y, \sqrt{3}y, and 2y.  So

2y=x=10    

Zigmanuir [339]3 years ago
5 0

Answer: A

Step-by-step explanation:

The line 'd' separates the square into two triangles. Since the side lengths are equal, the angle is 45°. We need the line separating the two 90° angles, that I am going to define as 'z'. We can use the sine function.

\sin(45)=\frac{z}{5\sqrt{2} }

z = 5

To solve for x, we can use sine again but for the 30 degree angle.

\sin(30)=\frac{5}{x} \\x=10

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Hi,so I've tried solving this problem for a while and I just don't get it,if you could give me some ideas I would appreciate it.
ziro4ka [17]

Answer:

187 cm²

Step-by-step explanation:

The bottom rectangle area is easy, it is 15*9 = 135 cm².

To find the area of the triangle, you only need the base width and its height (you don't need the hypotenuse). You can then use the formula: area triangle is base times half height.

The base width is 15-7 = 8 cm

The height is 22-9 = 13 cm

So the area of the triangle is 13*8/2 = 52 cm²

Together with the 135 of the rectangle that sums to 52+135 = 187 cm².

4 0
3 years ago
Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s
Snezhnost [94]

Answer. First option: t > 6.25


Solution:

Height (in feet): h=-16t^2+729

For which interval of time is h less than 104 feet above the ground?

h < 104

Replacing h for -16t^2+729

-16t^2+729 < 104

Solving for h: Subtracting 729 both sides of the inequality:

-16t^2+729-729 < 104-729

-16t^2 < -625

Multiplying the inequality by -1:

(-1)(-16t^2 < -625)

16t^2 > 625

Dividing both sides of the inequality by 16:

16t^2/16 > 625/16

t^2 > 39.0625

Replacing t^2 by [ Absolute value (t) ]^2:

[ Absolute value (t) ]^2 > 39.0625

Square root both sides of the inequality:

sqrt { [ Absolute value (t) ]^2 } > sqrt (39.0625)

Absolute value (t) > 6.25

t < -6.25 or t > 6.25, but t can not be negative, then the solution is:

t > 6.25



5 0
3 years ago
Read 2 more answers
Determine whether each expression can be used to find the length of side AB. Match Yes or No for each
tankabanditka [31]

Answer:

(a)\ AB = \frac{7}{\sin (B)}  \to Yes

(b)\ AB = \frac{24}{\cos (B)} \to Yes

(c)\ AB = \frac{24}{\cos (A)} \to No

(d)\ AB = \frac{7}{\cos (A)}  \to Yes

Step-by-step explanation:

Given

BC =24

AC = 7

Required

Select Yes or No for the given options

(a)\ AB = \frac{7}{\sin (B)}  \to Yes

Considering the sine of angle B, we have:

\sin(B) = \frac{Opposite}{Hypotenuse}

\sin(B) = \frac{7}{AB}

Make AB, the subject

AB = \frac{7}{\sin(B)}

(b)\ AB = \frac{24}{\cos (B)} \to Yes

Considering the cosine of angle B, we have:

\cos(B) = \frac{Adjacent}{Hypotenuse}

\cos(B) = \frac{24}{AB}

Make AB the subject

AB = \frac{24}{\cos(B)}

(c)\ AB = \frac{24}{\cos (A)} \to No

Considering the cosine of angle B, we have:

\cos(A) = \frac{Adjacent}{Hypotenuse}

\cos(A) = \frac{7}{AB}

Make AB the subject

AB = \frac{7}{\cos(A)}

(d)\ AB = \frac{7}{\cos (A)}  \to Yes

<em>This has been shown in (c) above</em>

3 0
3 years ago
You are 25 dollars in debt. You borrow 12 dollars more. What is the total amount of your debt? Justify your reasoning.
juin [17]

Answer:

you would be 37 dollars in debt.

Step-by-step explanation:

If you already are 25 dollars in debt and borrow 12 more dollars you would add that 12 dollars to your already 25 dollar in debt which would mean you would be 37 dollars in debt.

4 0
4 years ago
Read 2 more answers
For which intervals the graphs of the functions f(x) = x^3 + x^2 - 4x - 4 is positive
Nookie1986 [14]

Step-by-step explanation:

Consider a function

f

(

x

)

which is twice differentiable. The graph of such a function will be concave upwards in the intervals where the second derivative is positive and the graph will be concave downwards in the intervals where the second derivative is negative. To find these intervals we need to find the inflection points i.e. the x-values where the second derivative is 0.

5 0
3 years ago
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