<u>T</u><u>h</u><u>e</u><u> </u><u>s</u><u>t</u><u>a</u><u>t</u><u>e</u><u>m</u><u>e</u><u>n</u><u>t</u><u>s</u><u> </u><u>t</u><u>r</u><u>u</u><u>e</u><u> </u><u>f</u><u>o</u><u>r</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>g</u><u>i</u><u>v</u><u>e</u><u>n</u><u> </u><u>f</u><u>u</u><u>n</u><u>c</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>a</u><u>r</u><u>e</u><u>:</u>
I tried them all and these two were correct, their solutions are as follows:
= f(x) = 1/2x + 3/2
= f(0) = 1/2 × 0 + 3/2
= f(0) = 0 + 3/2
= f(0) = 3/2
= f(x) = 1/2x + 3/2
= f(4) = 1/2 × 4 + 3/2
= f(4) = 2 + 3/2
= f(4) = 4+3/2
= f(4) = 7/2
So, that's how these two are correct.
Answer:212 seats
Step-by-step explanation:a1=60
D=8
Answer:
y=2/3x+14
Step-by-step explanation:
The standard form of an equation in slope-intercept form is y=mx+b where m=slope and b=y-intercept.
Given a y-intercept of 14 and a slope of 2/3, we can plug into the variables and get the equation y=2/3x+14
The x-intercept would be when y=0, so plugging in y=0 to the equation gets us:
0=2/3x+14
-14=2/3x
21=x
So the x-intercept is 21
H = 3b+2
A = (h*b)/2 28 = (3b+2)b/2 56 = 3b²+2b 0 = 3b² + 2b - 56
⊕
![\left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta \\ \\ \\ x^{2} \sqrt{x} \sqrt[n]{x} \frac{x}{y} x_{123} x^{123} \leq \geq \pi \alpha \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] x_{123} \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}}](https://tex.z-dn.net/?f=%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%5Cbeta%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20x%5E%7B2%7D%20%20%5Csqrt%7Bx%7D%20%20%5Csqrt%5Bn%5D%7Bx%7D%20%20%5Cfrac%7Bx%7D%7By%7D%20%20x_%7B123%7D%20%20x%5E%7B123%7D%20%20%5Cleq%20%20%5Cgeq%20%20%5Cpi%20%20%5Calpha%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20x_%7B123%7D%20%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D)
ω
l
∩
<u>Answer</u>
4m
<u>Explanation</u>
Δqrp is a right triangle.
If ps = 6m then, qp = 3m. So we can find line pr using pythagoras theorem.
rp = √(5²-3²) = √16 = 4m
The Δqrp≡Δqtp.