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Yuki888 [10]
2 years ago
7

Antonio and Candice had a race. Their distances past a particular landmark as the race progressed can be represented by linear f

unctions and are shown in the table of values and graph below. Which statement is true regarding Antonio and Candice’s race?
Antonio’s Distance Past the Landmark
Time (seconds)
8
12
16
20
24
Distance (meters)
29
41
53
65
77

A graph titled Candice's Distance Past the Landmark has time (seconds) on the x-axis, and Distance (meters) on the y-axis. A line goes through points (5, 22), (10, 42), (15, 62) and (20, 82).
Mathematics
2 answers:
Zolol [24]2 years ago
9 0

Answer:

<u>Antonio had a head start of 3 meters.</u>

Step-by-step explanation:

I took the test on Edge and got it correct.

<em>(Sorry, this is late and you probably don't need it anymore.)</em>

Novosadov [1.4K]2 years ago
6 0

Answer: 3 meter for Antionio

Step-by-step explanation:

Whoopie i got i 100 trust me.

WHHOP WHOOP

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Which graph shows the solution to the system of linear inequalities? y ≥ 2x + 1 y ≤ 2x – 2
Artyom0805 [142]

Answer:

Step-by-step explanation:

Given the two inequalities:

y \ge 2x + 1\\ y \le 2x - 2

To graph them, first of all, let us write their corresponding equations:

y = 2x + 1 \\y = 2x - 2

We now find at least 2 points each which satisfy the equations to plot the graph.

Equation y = 2x + 1:

Putting x = 0, y = 1

Putting y = 0, x = -\frac{1}2.

Two points are (0, 1) and (-\frac{1}2, 0).

Equation y = 2x -2:

Putting x = 0, y = -2

Putting y = 0, x = 1.

Two points are (0, -2) and (1, 0).

Now, let us plot them.

Let us take a point (1, 4) and check whether it satisfies the first inequality.

4 \ge 2 \times 1 +1\\\Rightarrow 4 \ge 3

Which is true.

So, The shaded region will be <em>towards point (1,4).</em>

<em></em>

Let us take a point (1, 4) and check whether it satisfies the second inequality.

4 \le 2 \times 1 -2\\\Rightarrow 4 \le 0\ [\bold{False}]

So, The shaded region will be <em>opposite to point (1,4).</em>

Please refer to the attached graph for the answer.

<em>No solution exists for them because there is no common shaded region</em>.

3 0
2 years ago
Rewrite in simplest radical form x 5/6 x 1/6
Natasha_Volkova [10]

Answer:

x^{\frac{5}{6}}/x^{\frac{1}{6}} = \sqrt[3]{x^2}

Step-by-step explanation:

Given

x^{\frac{5}{6}}/x^{\frac{1}{6}}

Required

Rewrite in simplest radical form

Using laws of indices:

a^m/a^n = a^{m-n}

This implies that

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{5}{6} - \frac{1}{6}}

Solve Exponents

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{5 - 1}{6} }

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{4}{6} }

Simplify exponent to lowest fraction

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{2}{3} }

Using laws of indices:

a^{\frac{m}{n}} = \sqrt[n]{a^m}

This implies that

x^{\frac{5}{6}}/x^{\frac{1}{6}} = \sqrt[3]{x^2}

This is as far as the expression can be simplified

3 0
2 years ago
Compute i^1+i^2+i^3....i^99+i^100
fgiga [73]

Good morning ☕️

Answer:

<h3>i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰ = 0</h3>

Step-by-step explanation:

Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰

S =  i¹ +  i² +  i³ + . . . + i⁹⁹  +  i¹⁰⁰

S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀

then, S is the sum of 100 consecutive terms of a geometric sequence (an)

where the first term a1 = i¹ = i  and the common ratio = i

FORMULA:______________________

S=(term1)*\frac{1-(common.ratio)^{number.of.terms}}{1-(common.ratio)}

_______________________________

then

S=i*\frac{1-i^{100} }{1-i}

or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1  (we know that i⁴ = 1)

Hence

S = 0

7 0
2 years ago
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SOVA2 [1]
The answer would be 1
Slope formula is y2-y1/x2-x1
8-2/8-2
6/6
1
3 0
2 years ago
Read 2 more answers
A farmer is estimating the surface area of his barn to find how much paint he needs to buy. One part of the barn is triangular a
Nady [450]
We know the triangle is an isosceles triangle. So, we use the Pythagorean Theorem. 

a^2+b^2=c^2

The c would be 22, because it's the longest edge. Now find a or b. They are the same, that is how you find the two sides of the triangle for part A.

To find the area of the triangle, the forumla is a=hb/2

Plug in the numbers for that formula and the answer would be found for part B.

I hope I helped.
7 0
2 years ago
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