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LenKa [72]
4 years ago
10

What is the solution of square root x-4+5=2 ? x = –17 x = 13 x = 53 no solution

Mathematics
1 answer:
professor190 [17]4 years ago
6 0
What is inside the square root? Is it just the root of x? Or is it ROOT(x-4) + 5 = 2? 
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Solve for y: 76y = 19
Lena [83]

Answer:

y = 1/4 = 0.250.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Learning Task 1: Identify similar and dissimilar fractions. On your note- book write S if the fractions are similar and D if dis
Ronch [10]
<h2><u>Complete Question: </u></h2>

Learning Task 1: Identify similar and dissimilar fractions. On your note- book write S if the fractions are similar and D if dissimilar.

1. \frac{2}{3} $ and $ \frac{1}{3}

2. \frac{3}{4} $ and $ \frac{1}4}

3. \frac{4}{7} $ and $ \frac{7}{8}

4. \frac{2}{5} $ and $ \frac{5}{11}

5. \frac{7}{13} $ and $ \frac{7}{9}

<h2><em><u>The answers:</u></em></h2>

1. \frac{2}{3} $ and $ \frac{1}{3} - Similar (S)

2. \frac{3}{4} $ and $ \frac{1}4} - Similar (S)

3. \frac{4}{7} $ and $ \frac{7}{8} - Dissimilar (D)

4. \frac{2}{5} $ and $ \frac{5}{11} - Dissimilar (D)

5. \frac{7}{13} $ and $ \frac{7}{9} - Dissimilar (D)

Note:

  • Similar fractions have the same denominator. i.e. the bottom value of both fractions are the same.
  • Dissimilar fractions have different value as denominator, i.e. the bottom value of both fractions are not the same.

Thus:

1. \frac{2}{3} $ and $ \frac{1}{3} - They have equal denominator. <u><em>Both fractions are similar (S).</em></u>

2. \frac{3}{4} $ and $ \frac{1}4} - They have equal denominator. <em><u>Both fractions are similar (S).</u></em>

3. \frac{4}{7} $ and $ \frac{7}{8} - They have equal denominator. <em><u>Both fractions are dissimilar (D).</u></em>

4. \frac{2}{5} $ and $ \frac{5}{11} - They have equal denominator. <u><em>Both fractions are dissimilar (D).</em></u>

5. \frac{7}{13} $ and $ \frac{7}{9} - They have equal denominator. <em><u>Both fractions are dissimilar (D).</u></em>

Therefore, the fractions in <em><u>1 and 2 are similar (S)</u></em> while those in <em><u>3, 4, and 5 are dissimilar (D).</u></em>

<em><u></u></em>

Learn more here:

brainly.com/question/22099172

7 0
3 years ago
Write these numbers in order of size, start with the<br> smallest number<br><br> 60% 1/2 0.3 3/4 0.4
adoni [48]
(Smallest to biggest) 0.3, 0.4, 1/2, 60%, 3/4
Turning all number to decimals
.3, .4, .5, .60, .75
8 0
3 years ago
Which quadratic function has a wider graph than y=2x^2?
Trava [24]

Answer:

y = 1/2 x²

Step-by-step explanation:

The coefficient of the first term in a quadratic, in our case here, x², will tell us how the graph stretches. This is akin to the slope within the linear graph. Similar to the slope, the smaller the coefficient value, or value of slope m, the shallower the angle.

When discussing quadratics, the larger the coefficient of our x² term, the steeper, and skinnier the graph. If we want to look for a graph that is wider than y = 2x², then we need to find a graph with a coefficient that is less than 2.

Our only option then is

y = 1/2 x²

8 0
3 years ago
Find the zeros of the function f(x) = x2 + 5x + 6. A) y = 6 because the graph crosses the y-axis at 6. B) y = -0.25 because that
Fofino [41]

Answer:

D.)

Step-by-step explanation:

The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts.  You can graph the function and see when the graph crosses the x-axis or solve for the x-values.  I will solve it via factoring and so:

f(x)=x^2+5x+6

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6.  Now let's think about all the factors of 6 we have: 6×1 and 2×3.  Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5?  Actually we can say 6-1=5 and 2+3=5.  Let's try both.

First let's use 6 and -1 and so:

x^2+5x+6\\\\x^2+6x-x+6\\\\x(x+6)-1(x-6)

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.

Now let's try 2 and 3 and so:

x^2+5x+6\\\\x^2+3x+2x+6\\\\x(x+3)+2(x+3)\\\\(x+2)(x+3)

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:

Case 1:

f(x)=x+2\\\\0=x+2\\\\x=-2

Case 2:

f(x)=x+3\\\\0=x+3\\\\x=-3

So your zero's are when x=-2 and x=-3.

D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.


~~~Brainliest Appreciated~~~

8 0
3 years ago
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