The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
</u>
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.
<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>
<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>
<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>
<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>
<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>
<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>
<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>
<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>
<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>
<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>
<span>Or, if you prefer, (3, 0), (-5, 0). </span>
Answer:
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Step-by-step explanation:
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We select a sample size n from the confidence interval with the mean 
and default 
, then the mean take seriously given as the straight line with a z score given by the confidence interval
Using formula:
The probability that perhaps the mean shells length of the sample is over 4.03 pounds is
Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution
the required probability:
