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3 years ago

the equation (x+7)(x-1)=-16 has a. two real solutions b. one real solution c. two complex solutions d. no solution

Mathematics

2 answers:

TEA [102]3 years ago

6 0

(x+7)(x-1) = -16

x² + 6x - 7 = -16

x² + 6x - 7 + 16 = 0

x² + 6x + 9 = 0

(x+3)² = 0

x = - 3 -> one real solution

Anni [7]3 years ago

5 0

Here’s my answer! Hope it makes sense :)

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saul85 [17]

The surface area of the geometric shape is B. 840 cm².

Step-by-step explanation:

Step 1:

The given figure is made up of three rectangles and two triangles.

The surface area of the geometric shape is given by summing the individual surface areas.

The geometric shape's surface area = The sum of the individual surface areas.

Step 2:

The surface area of the rectangles is the multiplication of its length and width.

The surface area of the rectangle with length z and width x $= (26)(10)= 260 cm^{2}$ .

The surface area of the rectangle with length y and width x $= (24)(10)= 240 cm^{2}$ .

The surface area of the rectangle with length x and width x $= (10)(10)= 100 cm^{2}$ .

Step 3:

The area of a triangle is half the product of its base length and height.

The surface area of a triangle with base y and height x $= \frac{1}{2} (24)(10) = 120 cm^{2}$ .

They are two similar triangles so their areas are equal

Step 4:

The geometric shape's surface area $= 160 + 240+100+120+120 = 840 cm^{2}$ .

So the surface area is option B. 840 cm².

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3 years ago

A cafeteria manager needs to know how many apples and bananas to to order. He asked students to choose either an apple or a bana

Usimov [2.4K]

Answer:

Step-by-step explanation:

3/4 of total = 45

therefore, 1/4 took bananas

the ones taking bananas are 3 times less than those who took apples.

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3 years ago

Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.

solong [7]

Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is: $f(x)= \frac{9-x^2}{x^2-4}$

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

a. (-infinity, -3)

put k = -4 from the given interval

We get $f(-4)= \frac{9-(-4)^2}{(-4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get $f(-2.5)= \frac{9-(-2.5)^2}{(-2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

We get $f(0)= \frac{9-(0)^2}{(0)^2-4} = \frac{9}{-4} = -\frac{9}{4} < 0$

⇒ f (x) < 0 for x ∈ (-2,2)

d. (2, 3)

put k =2.5 from the given interval

We get $f(2.5)= \frac{9-(2.5)^2}{(2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (2,3)

e. (infinity, 3)

put k = 4 from the given interval

We get $f(4)= \frac{9-(4)^2}{(4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (3,∞)

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4 years ago

Help me please, ill will mark you brainliest

masha68 [24]

Answer:

Step-by-step explanation:

c=37

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4 years ago

Which expression is equivalent? (2−1x−3y−3)5

irga5000 [103]

−5x−15y−5 is what i get

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3 years ago