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Marat540 [252]
3 years ago
12

A. Consider the cube shown below. Identify the two-dimensional shape of the cross-section if the cube is sliced horizontally.

Mathematics
1 answer:
irina [24]3 years ago
8 0

Answer:

300

Step-by-step explanation:

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If tan A = 4/3 and sin B = 45/53 and angles A and B are in Quadrant I, find the value of tan(A+B)
lbvjy [14]

Answer:

First, find tan A and tan B.

cosA=35 --> sin2A=1−925=1625 --> cosA=±45

cosA=45 because A is in Quadrant I

tanA=sinAcosA=(45)(53)=43.

sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.

sinB=1213 because B is in Quadrant I

tanB=sinBcosB=(513)(1312)=512

Apply the trig identity:

tan(A−B)=tanA−tanB1−tanA.tanB

tanA−tanB=43−512=1112

(1−tanA.tanB)=1−2036=1636=49

tan(A−B)=(1112)(94)=3316

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3 0
2 years ago
(1/4)^3z-1 =16^z+2*64^z-2<br><br> Z=___<br><br> Please help
wariber [46]

Answer:

Z = 0.198877274

Step-by-step explanation:

(\frac{1}{4})^{3z-1} = 16^z + 2*16^{z-2}\\4^{1-3z} = 4^{2z} + 4^{\frac{1}{2}}*4^{2z-4}\\4^{1-3z} = 4^{2z} + 4^{2z-4+\frac{1}{2}}\\4^{1-3z} = 4^{2z} + 4^{2z-\frac{7}{2}}\\4^{1-3z} = 4^{2z} *(1+ 4^{-\frac{7}{2}})\\4^{1-3z} = 4^{2z} *(1+ 2^{-7})\\4^{1-3z} = 4^{2z} *(1+ \frac{1}{128} )\\4^{1-3z} = 4^{2z} *(\frac{129}{128} )\\Taking\;\; Logarithm\;\; with\;\; base\;\; 4\\Log_4(4^{1-3z}) = Log_4(4^{2z}) + Log_4(\frac{129}{128})\\1-3z = 2z + 0.005613627712 \\5z = 0.994386372\\z = 0.198877274

Hence, the value of Z = 0.198877274

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3 years ago
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The length of a rectangular floor is 2 feet more than its width. The area of the floor is 168 square feet. Kim wants to use a ru
ioda
Let width of the floor be x feet, them area is
x(x+2)  = 168
x^2 + 2x - 168 = 0
(x - 12)(x + 14) - 0
so x = 12
the floor is 12 ft by 10 ft

the width of the rug should be 10 - 2(2) = 6 feet


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