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3 years ago

The greatestcommonfactor of 8, 52

1 answer:

Nadusha1986 [10]3 years ago

6 0

The GCF is the largest number that both will divide into. Eight is the larger of the two, but 52 is not divisible by 8. The next highest value is 4. 8÷4=2 and 52÷4=13. Therefore, the GCF is 4.

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If the original point A (3,5) is rotated 180 degrees what will the image point be

Grace [21]

Answer:

5,3 I think or 5,4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A can of soda is placed inside a cooler. As the soda cools, its temperature T(x) in degrees Celsius is given by the following ex

Lana71 [14]

Answer:

Ans: -21.87 ≅ -22°C

Step-by-step explanation:

T(x)=-22+44e^-0.03x

Initial temperature (x = 0):

T(0) = = -22 + 44e-0.03(0) = -22 + 44(1) = 22°C

After 18 minutes (x = 18):

T(18) = -22 + 44e-0.03(18) = -21.87°C ≅ -22°C

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2 years ago

Draw the image of the following segment after a dilation centered at the origin with a scale factor of 2/3

Irina18 [472]

Answer:

Step-by-step explanation:

Let the ends of the given segment are A and B.

Coordinates of A → (8, 6)

Coordinates of B → (12, 12)

If a point (x, y) is dilated by a scale factor 'k' about the origin, rule to be followed,

(x, y) → (kx, ky)

If k = $\frac{2}{3}$

(x, y) → $(\frac{2}{3}x, \frac{2}{3}y)$

By this rule coordinates of the image points of A and B will be,

A(8, 6) → $A'(\frac{2\times 8}{3},\frac{2\times 6}{3})$

→ A'(5.3, 4)

B(12, 12) → $B'(\frac{12\times 2}{3}, \frac{12\times 2}{3})$

→ B'(8, 8)

Now we can get the image of segment AB after dilation by a scale factor of $\frac{2}{3}$ .

5 0

2 years ago

In ABC, BC = a = 16, AC = b = 10, and m<C = 22º. Which equation can you use to find the value of c = AB?

AnnyKZ [126]

Answer:

10 times 16 = C

Step-by-step explanation:

c2 = a2 + b2 − 2ab cos C = 162 + 102 − 2(10)(16) cos 22°

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3 years ago

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.<br>

Lemur [1.5K]

In both cases,

$AB^2=AD^2$

(as a consequence of the interesecting secant-tangent theorem)

So we have

10.

$(4x+7)^2=(6x-3)^2$

$16x^2+56x+49=36x^2-36x+9$

$20x^2-92x-40=0$

$5x^2-23x-10=0$

$(5x+2)(x-5)=0\implies\boxed{x=5}$

(omit the negative solution because that would make at least one of AB or AD have negative length)

11.

$(4x^2-18x-10)^2=(x^2+x+4)^2$

$16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16$

$15x^4-146x^3+235x^2+352x+84=0$

$(x-7)(3x+2)(5x^2-17x-6)=0\implies\boxed{x=-\dfrac23\text{ or }x=7}$

(again, omit the solutions that would give a negative length for either AB or AD)

7 0

3 years ago