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Sloan [31]
3 years ago
9

Which of the following is the equation for the graph shown?

Mathematics
1 answer:
photoshop1234 [79]3 years ago
3 0

Answer:

x^2/25-y^/16=1 or y^2/16-x^2/25=1

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Which is greater 0.8 or 3/4
Illusion [34]
3/4 is equivalent to 0.75  Therefore 0.8 is greater
7 0
3 years ago
Read 2 more answers
What is the answer to 72% of 14?
SSSSS [86.1K]
14/100 = 0.14
0.14x72 = 10.08
7 0
3 years ago
X-3 and 2x-1 are factors of 2x^3-px^2-2qx+q. Find the values of p and q.
Nonamiya [84]

Answer:

p =1

q = 9

Step-by-step explanation:

f(x) = 2x³ - px² + 2qx + q

(x - 3) is a factor of f(x)

⇒f(3)  = 0

2(3)³ - p*3² - 2q*3 +q = 0

2*27 - 9p - 6q + q = 0

    54 - 9p - 5q = 0

            -9p - 5q = -54 -------------------(I)

(2x - 1) is a factor of f(x)

2x - 1 = 0

      2x = 1

        \st x = \dfrac{1}{2}

f(1/2) = 0

2*(\dfrac{1}{2})^{3}-p*(\dfrac{1}{2})^{2}-2q*\dfrac{1}{2}+q=0\\\\2*\dfrac{1}{8}-p*\dfrac{1}{4}-q+q = 0\\\\\dfrac{1}{4}-\dfrac{1}{4}p =0\\\\[Multiply the entire equation by 4]\\\\4*\dfrac{1}{4}-4*\dfrac{1}{4}p=0\\\\

1 - p = 0

    -p = -1

p = 1

Substitute p =1 in equation (I)

-9*1 - 5q = -54

 -9 - 5q = -54

      -5q = -54 + 9

      -5q = -45

           q = -45/(-5)

q = 9

8 0
3 years ago
Dingane has $8.00, and exactly 30% of that money is from 5-cent coins. how many 5-cent coins does dingane have?
il63 [147K]
Answer=48

30%=0.30

$8.00*0.30=$2.40

Dingane has $2.40 worth of 5-cent coins

$2.40/0.05=48

He has 48 5-cent coins
6 0
4 years ago
Can anyone do 4,5 and 6 for me plz
cricket20 [7]

For  question 4, sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.

Question 5. Option a and question 6. Option j

Step-by-step explanation:

Step 1:

The three basic formula needed to solve these questions are:

sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.

Step 2:

Using the above formula, we solve the following values

sinA = \frac{oppositeside}{hypotenuse}  =\frac{a}{c}.

cosA = \frac{adjacentside}{hypotenuse} = \frac{b}{c}.

tanB= \frac{opposite side}{adjacent side}= \frac{b}{a}.

sinJ = \frac{oppositeside}{hypotenuse} = \frac{j}{l}.

cosK = \frac{adjacentside}{hypotenuse}= \frac{j}{l}.

tanK= \frac{opposite side}{adjacent side}= \frac{k}{j}.

Step 3:

For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.

sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.

SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.

Step 4:

For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.

tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277

SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.

7 0
3 years ago
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